JEE Ray Optics Questions

The paper is burnt fastest when the Sun (an object at infinity) is imaged as the smallest, brightest spot on the paper. For a thin lens, an object at infinity is brought to focus at the focal plane. Hence the distance between the lens and the paper equals the focal length of the lens.

Given distance between lens and paper: $$f = 30 \text{ cm}$$.

The lens is symmetric and convex with radius of curvature $$R = 60 \text{ cm}$$ on both faces. Using the lens-maker’s formula for a thin lens in air:

$$\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

For a double-convex lens:
  • First surface: centre of curvature on the right ⇒ $$R_1 = +60 \text{ cm}$$.
  • Second surface: centre of curvature on the left ⇒ $$R_2 = -60 \text{ cm}$$.

Substituting these values:

$$\frac{1}{30} = (n-1)\left(\frac{1}{60} - \left(-\frac{1}{60}\right)\right)$$

$$\frac{1}{30} = (n-1)\left(\frac{2}{60}\right) = (n-1)\left(\frac{1}{30}\right)$$

Multiplying both sides by $$30$$:

$$1 = n - 1 \quad \Longrightarrow \quad n = 2$$

The refractive index is written as $$n = \dfrac{\alpha}{10}$$, so

$$\alpha = 10n = 10 \times 2 = 20$$

Hence, the required value of $$\alpha$$ is 20.

Let the spherical surface have its vertex at point $$V$$ on the optic axis, with the centre of curvature $$C$$ lying inside the glass medium at a distance $$R$$ from $$V$$.
Choose the Cartesian sign convention: light travels from left (air) to right (glass).
Hence, distances measured to the right of $$V$$ are taken as positive.

Given data
  • Refractive index in air, $$n_1 = 1$$
  • Refractive index in glass, $$n_2 = 1.5$$
  • Radius of curvature, $$R$$, is positive (centre is on the image side).

The object $$O$$ is in air (left side), so its distance from the vertex is $$u = -PO$$ (negative).
The real image $$I$$ is in glass (right side), so its distance from the vertex is $$v = PI$$ (positive).
It is given that $$PO = PI$$, therefore the magnitudes of the object and image distances are equal:
  $$|u| = v \quad\Longrightarrow\quad v = -u$$ $$-(1)$$

The refraction at a spherical surface is governed by the formula
$$\frac{n_2}{v} - \frac{n_1}{u} = \frac{n_2 - n_1}{R}$$ $$-(2)$$

Substitute $$n_1 = 1$$, $$n_2 = 1.5$$ and use $$(1)$$, i.e. $$u = -v$$:

$$\frac{1.5}{v} - \frac{1}{-v} = \frac{1.5 - 1}{R}$$

Simplify the left-hand side:
$$\frac{1.5}{v} + \frac{1}{v} = \frac{2.5}{v}$$

Thus
$$\frac{2.5}{v} = \frac{0.5}{R}$$

Solving for $$v$$:
$$v = \frac{2.5}{0.5}\,R = 5R$$

The distance asked in the problem is $$PO$$, and $$PO = v = 5R$$.

Therefore, the correct option is Option A: $$5R$$.

$$|m| = 2$$ (Hence, mirror is concave)

$$\text{Distance between object and image} = 15\text{ cm}$$

For a virtual image in a mirror, the image is formed behind the mirror while the object is in front:

$$\text{Distance} = |v| + |u| = 15\text{ cm}$$

Magnification relation for mirrors: $$m = -\frac{v}{u} \implies |v| = |m||u| = 2|u|$$

$$2|u| + |u| = 15 \implies 3|u| = 15 \implies |u| = 5\text{ cm}$$

$$|v| = 2 \times 5 = 10\text{ cm}$$

$$u = -5\text{ cm},\quad v = +10\text{ cm}$$

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \implies \frac{1}{f} = \frac{1}{10} + \frac{1}{-5} = -\frac{1}{10} \implies f = -10\text{ cm}$$

For lenses in contact, the effective power is the sum of individual powers:

$$P_{total} = 5P$$ where P is the power of each lens.

$$25 = 5P \Rightarrow P = 5$$ D

Focal length: $$f = \frac{1}{P} = \frac{1}{5}$$ m = 20 cm

The correct answer is Option 1: 20 cm.

Using the power of a combined silvered thin lens system:

$$\frac{1}{F} = 2\left(\frac{n - 1}{R_1} + \frac{\mu_g - n}{R_2}\right) + \frac{1}{f_m}$$

Given parameters: flat air-liquid interface ($$R_1 = \infty$$), liquid-glass convex interface ($$R_2 = +9\text{ cm}$$), and flat mirror base ($$f_m = \infty$$):

$$\frac{1}{F} = 2\left(0 + \frac{1.60 - n}{9}\right) + 0 = \frac{2(1.60 - n)}{9}$$

Using the coincidence condition where the image reflects back to the object position ($$u = -h$$, $$v = -h$$):

$$\frac{1}{-h} + \frac{1}{-h} = -\frac{1}{F} \implies \frac{2}{h} = \frac{2(1.60 - n)}{9}$$

$$h = \frac{9}{1.60 - n}$$

For Option (A): $$n = 1.42 \implies h = \frac{9}{1.60 - 1.42} = \frac{9}{0.18} = 50\text{ cm}$$

For Option (B): $$n = 1.35 \implies h = \frac{9}{1.60 - 1.35} = \frac{9}{0.25} = 36\text{ cm}$$

For Option (C): $$n = 1.45 \implies h = \frac{9}{1.60 - 1.45} = \frac{9}{0.15} = 60\text{ cm}$$

For Option (D): $$n = 1.48 \implies h = \frac{9}{1.60 - 1.48} = \frac{9}{0.12} = 75\text{ cm}$$

For each equilateral-triangular prism the refracting (apex) angle is $$A = 60^\circ = \dfrac{\pi}{3}$$.

Step 1: Condition for minimum deviation in prism $$P_2$$
For minimum deviation in any prism, the path of the ray is symmetrical, so $$r = r' = \dfrac{A}{2} = 30^\circ$$ inside the prism, and the angles of incidence and emergence are equal: $$i = e$$.

Applying Snell’s law at the first face of $$P_2$$ (outside medium is vacuum, $$n = 1$$):
$$\sin i = n_2 \sin r$$

Given $$n_2 = \sqrt{3}$$, therefore
$$\sin i = \sqrt{3}\,\sin 30^\circ = \sqrt{3}\left(\dfrac{1}{2}\right)=\dfrac{\sqrt{3}}{2}$$
so $$i = 60^\circ = \dfrac{\pi}{3}$$.

Hence the light ray must strike the first face of $$P_2$$ with an incidence angle of $$60^\circ$$.

Step 2: This same ray is the emergent ray from prism $$P_1$$
The two prisms are arranged with their corresponding faces parallel; therefore the angle the ray makes with the normal when it leaves $$P_1$$ is the same $$60^\circ$$. Let that angle of emergence from $$P_1$$ be $$e_1 = 60^\circ$$.

Step 3: Ray path inside prism $$P_1$$
Let the angles of refraction at the two faces of $$P_1$$ be $$r_1$$ and $$r_2$$. For any prism
$$r_1 + r_2 = A = 60^\circ \qquad -(1)$$

Snell’s law at the second face (emergence) of $$P_1$$:
$$n_1 \sin r_2 = \sin e_1$$
Given $$e_1 = 60^\circ$$ and $$n_1 = \sqrt{\dfrac{3}{2}}$$, we get
$$\sin r_2 = \dfrac{\sin 60^\circ}{n_1} = \dfrac{\dfrac{\sqrt{3}}{2}}{\sqrt{\dfrac{3}{2}}} = \dfrac{\sqrt{2}}{2} = \sin 45^\circ$$
so $$r_2 = 45^\circ$$.

Using $$(1)$$:
$$r_1 = 60^\circ - 45^\circ = 15^\circ.$$

Step 4: Incidence angle $$\theta$$ on the first face of $$P_1$$
Apply Snell’s law at the first face of $$P_1$$:
$$\sin \theta = n_1 \sin r_1 = \sqrt{\dfrac{3}{2}}\;\sin 15^\circ$$

Since $$15^\circ = \dfrac{\pi}{12}$$,
$$\sin \theta = \sqrt{\dfrac{3}{2}}\;\sin\!\left(\dfrac{\pi}{12}\right).$$

The question states this result as $$\theta = \sin^{-1}\!\left[\sqrt{\dfrac{3}{2}}\;\sin\!\left(\dfrac{\pi}{\beta}\right)\right].$$ Comparing, we must have $$\dfrac{\pi}{\beta} = \dfrac{\pi}{12}\quad\Longrightarrow\quad \beta = 12.$$

Hence the required value is 12.

A microscope focused on an object at the bottom of a bucket must be raised 30 cm when a liquid of refractive index 5/3 is poured in. We need the height of the liquid.

Understand apparent depth and real depth.

When light travels from a denser medium (liquid) to a rarer medium (air), the apparent position of an object is closer to the surface than its actual position. The relationship is:

$$ \text{Apparent depth} = \frac{\text{Real depth}}{n} $$

where $$n$$ is the refractive index of the liquid.

Calculate the apparent shift.

The apparent shift (the amount the object appears to rise) is:

$$ \text{Shift} = \text{Real depth} - \text{Apparent depth} = h - \frac{h}{n} = h\left(1 - \frac{1}{n}\right) $$

This is the distance the microscope must be raised to refocus.

Substitute and solve.

Given: shift = 30 cm, $$n = 5/3$$:

$$ 30 = h\left(1 - \frac{1}{5/3}\right) = h\left(1 - \frac{3}{5}\right) = h \times \frac{2}{5} $$

$$ h = \frac{30 \times 5}{2} = 75\;\text{cm} $$

The height of the liquid is 75 cm.

The correct answer is Option 1: 75 cm.

The least distance of distinct vision (near point) is obtained from the data of the reading (converging) glasses, because these spectacles are chosen so that an object kept at the ordinary reading distance appears at the patient’s near point.

Step 1: Write the basic formulas
Power: $$P = \frac{1}{f}$$  (with $$f$$ in metres)
Lens formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$ $$-(1)$$
Here $$u$$ = object distance from the lens, $$v$$ = image distance from the lens (measured from the same side as the light is incident; a virtual image has negative $$v$$).

Step 2: Focal length of the reading glass
Given power of reading glass, $$P_r = +2.0\ \text{D}$$.
Therefore $$f_r = \frac{1}{P_r} = \frac{1}{2.0}\ \text{m} = 0.50\ \text{m}$$.

Step 3: Choose the usual reading distance
While reading, a book is normally held at about $$25\ \text{cm}$$ from the spectacles.
Hence for the lens, $$u = -25\ \text{cm} = -0.25\ \text{m}$$ (negative by the Cartesian sign convention).

Step 4: Locate the virtual image (near point) using Eq. $$(1)$$
Substituting $$f = +0.50\ \text{m}$$ and $$u = -0.25\ \text{m}$$:
$$\frac{1}{0.50} = \frac{1}{v} - \left(-\frac{1}{0.25}\right)$$
$$2 = \frac{1}{v} + 4$$
$$\frac{1}{v} = 2 - 4 = -2$$
$$v = -0.50\ \text{m}$$.

The negative sign confirms that the image is virtual and lies on the same side as the object (towards the eye). The magnitude gives the distance of the near point:

$$|v| = 0.50\ \text{m} = 50\ \text{cm}$$.

Step 5: State the result
The person’s least distance of distinct vision is $$50\ \text{cm}$$.

Therefore, the correct option is Option D (50 cm).

We need to determine what improves the resolving power of a compound microscope.

The resolving power formula:

The resolving power (RP) of a microscope is given by:

$$RP = \frac{2\mu \sin\theta}{\lambda}$$

where:

- $$\mu$$ is the refractive index of the medium between the object and the objective lens

- $$\theta$$ is the half-angle of the maximum cone of light entering the objective

- $$\lambda$$ is the wavelength of light used

The term $$\mu \sin\theta$$ is called the numerical aperture (NA) of the objective lens.

Analyzing each option:

Option A: Increase the wavelength of light — From the formula, RP is inversely proportional to $$\lambda$$. Increasing wavelength would decrease the resolving power. This is incorrect.

Option B: Increase the refractive index of the medium between the object and objective lens — From the formula, RP is directly proportional to $$\mu$$. Increasing $$\mu$$ (for example, by using oil immersion instead of air) would increase the resolving power. This is correct.

Option C: Decrease the focal length of the eyepiece — The focal length of the eyepiece affects the magnifying power of the microscope, not its resolving power. Resolving power depends on the objective lens and the medium, not the eyepiece. This is incorrect.

Option D: Decrease the diameter of the objective lens — A smaller objective lens aperture would reduce $$\sin\theta$$, thereby decreasing the resolving power. This is incorrect.

The correct answer is Option B: Increase the refractive index of the medium between the object and objective lens.

$$d_{\text{real1}} = \mu \cdot d_{\text{apparent1}}, \quad d_{\text{real2}} = \mu \cdot d_{\text{apparent2}}$$

$$\text{Total real side length: } L = d_{\text{real1}} + d_{\text{real2}} = \mu(d_{\text{apparent1}} + d_{\text{apparent2}})$$

$$24 = \mu(12 + 4) \implies 24 = 16\mu \implies \mu = \frac{24}{16} = \frac{3}{2}$$

Statement I: Brewster's angle air→glass: $$\tan\theta_B = \mu_g$$. Glass→air: $$\tan\theta'_B = 1/\mu_g = \cot\theta_B = \tan(\pi/2-\theta_B)$$. So $$\theta'_B = \pi/2-\theta_B$$. TRUE.

Statement II: Brewster's angle glass→air should be $$\tan^{-1}(1/\mu_g)$$, not $$\tan^{-1}(\mu_g)$$. FALSE.

Statement I is true but Statement II is false.

In a reflecting telescope, a secondary mirror is used to redirect the light collected by the primary concave mirror.

Understand the structure of a reflecting telescope. A reflecting telescope uses a large concave mirror (primary mirror) to collect and focus light. However, the focal point of the primary mirror lies in front of the mirror, which means the observer's head would block incoming light if the eyepiece were placed there.

Role of the secondary mirror. The secondary mirror (a small flat or convex mirror) is placed near the focal point of the primary mirror to redirect the light to a more convenient location. In a Cassegrain telescope, it reflects light back through a hole in the primary mirror. In a Newtonian telescope, it deflects light to the side of the tube.

Evaluate the options. Option A: "Reduce the problem of mechanical support" — The secondary mirror helps avoid the need to place heavy equipment at the prime focus, reducing mechanical support issues. This is correct.

Option B: "Make chromatic aberration zero" — Reflecting telescopes already have zero chromatic aberration (mirrors don't cause dispersion). The secondary mirror doesn't contribute to this.

Option C: "Move the eyepiece outside the telescopic tube" — This is also a benefit of the secondary mirror.

Option D: "Remove spherical aberration" — The secondary mirror doesn't remove spherical aberration.

The primary purpose is to reduce the mechanical support problem by redirecting light so equipment doesn't need to be at the prime focus.

The correct answer is Option A.

The critical angle $$\theta_c$$ for a denser-rarer interface is related to the refractive indices by:

$$\sin \theta_c = \frac{n_r}{n_d} = \frac{v_d}{v_r}$$

where $$v_d$$ and $$v_r$$ are the speeds of light in the denser and rarer media respectively.

We are given that $$\theta_c = 45°$$, $$v_r = 3 \times 10^8$$ m/s

$$\sin 45° = \frac{v_d}{v_r}$$

$$\frac{1}{\sqrt{2}} = \frac{v_d}{3 \times 10^8}$$

$$v_d = \frac{3 \times 10^8}{\sqrt{2}} = \frac{3 \times 10^8 \times \sqrt{2}}{2} = \frac{3 \times 1.414 \times 10^8}{2}$$

$$v_d = \frac{4.243 \times 10^8}{2} = 2.12 \times 10^8 \text{ m/s}$$

The speed of light in the denser medium is $$2.12 \times 10^8$$ m/s.

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies v = \frac{uf}{u - f}$$

$$v_A = \frac{u_A f}{u_A - f}$$

$$v_A = \frac{(-15)(-20)}{-15 - (-20)}$$

$$v_A = \frac{300}{-15 + 20}$$

$$v_A = \frac{300}{5} = +60 \text{ cm}$$

$$v_B = \frac{u_B f}{u_B - f}$$

$$v_B = \frac{(-25)(-20)}{-25 - (-20)}$$

$$v_B = \frac{500}{-5} = -100 \text{ cm}$$

$$d = |v_A - v_B| = |60 - (-100)| = 160 \text{ cm}$$

$$\text{By the law of reflection, the angle of incidence equals the angle of reflection: } i = r = 30^\circ$$

$$\delta = 180^\circ - (i + r)$$

$$\delta = 180^\circ - (30^\circ + 30^\circ) = 180^\circ - 60^\circ = 120^\circ$$

A plane mirror forms a virtual and erect image of the same size as the object. The image is also laterally inverted (left and right are interchanged).

Therefore:

• A. Real → Incorrect
• B. Erect → Correct
• C. Smaller in size than object → Incorrect
• D. Laterally inverted → Correct

For reflection from $$M_2$$:

$$\frac{1}{v} + \frac{1}{(-10)} = \frac{1}{(-12)}$$

$$\frac{1}{v} = \frac{1}{10} - \frac{1}{12}$$

$$v = +60\text{ cm}$$ (for $$I_1$$)

For refraction from $$L$$:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} - \frac{1}{(-80)} = \frac{1}{10}$$

$$v = +\frac{80}{7}$$ (For $$I_2$$)

This image should be at focus of $$\text{M}_1$$ $$\implies$$ $$\therefore \frac{20}{2} + \frac{80}{7} = \frac{n}{7}$$ $$\implies$$ $$n = 150$$

Also, If $$I_2$$ is formed at pole of $$M_1$$ then $$\frac{n}{7} = \frac{80}{7}$$ $$\implies$$ $$n = 80$$

And further if $$\text{I}_2$$ is formed at centre of curvature of $$\text{M}_1$$ then $$\frac{n}{7} = \frac{80}{7} + 20$$ $$\implies$$ $$\therefore n = 220$$

Solution :

For a converging mirror,

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$

Given :

$$u = 40\text{ cm}$$

$$v = 120\text{ cm}$$

Therefore,

$$\frac{1}{f} = \frac{1}{120} + \frac{1}{40}$$

$$= \frac{1+3}{120}$$

$$= \frac{1}{30}$$

$$f = 30\text{ cm}$$

Modified scale has :

$$20\text{ divisions} = 1\text{ cm}$$

Least count :

$$LC = \frac{1}{20}\text{ cm}$$

$$= 0.05\text{ cm}$$

Hence,

$$\Delta u = \Delta v = 0.05\text{ cm}$$

Using error propagation formula :

$$\frac{\Delta f}{f^2} = \frac{\Delta v}{v^2} + \frac{\Delta u}{u^2}$$

Substituting values :

$$\Delta f = (30)^2\left(\frac{0.05}{(120)^2} + \frac{0.05}{(40)^2}\right)$$

$$= 900 \times 0.05 \left(\frac{1}{14400} + \frac{1}{1600}\right)$$

$$= 45\left(\frac{1+9}{14400}\right)$$

$$= 45 \times \frac{10}{14400}$$

$$= \frac{450}{14400}$$

$$= \frac{1}{32}\text{ cm}$$

Given,

$$\Delta f = \frac{1}{K}\text{ cm}$$

Therefore,

$$K = 32$$

Final Answer :

$$32$$

Using apparent position relation: $$y_{\text{apparent}} = \mu y$$

Total apparent distance observed by fish: $$s = x + \mu y$$

Differentiating with respect to time for velocities:

$$\frac{ds}{dt} = \frac{dx}{dt} + \mu \frac{dy}{dt}$$

$$12 = 8 + \frac{4}{3} v_{\text{bird}}$$

$$4 = \frac{4}{3} v_{\text{bird}} \implies v_{\text{bird}} = 3\text{ m s}^{-1}$$

A biconvex lens of focal length 10 cm is cut perpendicular to the principal axis into two identical halves. Each half becomes a plano-convex lens.

For the original biconvex lens (assuming symmetric with equal radii of curvature $$R$$):

$$\frac{1}{f} = (n-1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (n-1)\left(\frac{2}{R}\right)$$

When cut perpendicular to the principal axis, each half is a plano-convex lens with one curved surface (radius $$R$$) and one flat surface (radius $$\infty$$):

$$\frac{1}{f'} = (n-1)\left(\frac{1}{R}\right) = \frac{1}{2} \times \frac{1}{f} = \frac{1}{2 \times 10} = \frac{1}{20}$$

$$f' = 20$$ cm $$= 0.2$$ m

Power: $$P = \frac{1}{f'} = \frac{1}{0.2} = 5$$ D

The power of each lens is $$\mathbf{5}$$ D.

$$\frac{1}{f} = \left(\frac{\mu_{\text{lens}}}{\mu_{\text{medium}}} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

$$\frac{1}{f_a} = (\mu_l - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

$$\frac{1}{f_w} = \left(\frac{\mu_l}{\mu_w} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

$$\frac{f_w}{f_a} = \frac{\mu_l - 1}{\frac{\mu_l}{\mu_w} - 1}$$

$$\frac{f_w}{f_a} = \frac{\frac{3}{2} - 1}{\frac{3/2}{4/3} - 1}$$

$$\frac{f_w}{f_a} = 4$$

$$f_w = 4 \times 35 = 140\text{ cm}$$

$$\Delta f = 140 - 35$$

$$\Delta f = 105\text{ cm}$$

We have two coaxial convex lenses: $$L_1$$ with focal length $$f_1 = 24$$ cm and $$L_2$$ with focal length $$f_2 = 9$$ cm, separated by 10 cm. An object is placed 6 cm from $$L_1$$, and we seek the distance between the object and the final image.

We apply the lens formula $$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$$ to the first lens, $$L_1$$. Here the object distance is $$u_1 = -6$$ cm (using the sign convention that left of the lens is negative) and the focal length is $$f_1 = 24$$ cm. Substituting these values gives $$\dfrac{1}{v_1} - \dfrac{1}{-6} = \dfrac{1}{24}$$, which simplifies to $$\dfrac{1}{v_1} + \dfrac{1}{6} = \dfrac{1}{24}$$. Therefore, $$\dfrac{1}{v_1} = \dfrac{1}{24} - \dfrac{1}{6} = \dfrac{1-4}{24} = \dfrac{-3}{24} = \dfrac{-1}{8}$$, giving $$v_1 = -8$$ cm. Hence the image formed by $$L_1$$ is 8 cm to the left of $$L_1$$ (a virtual image on the same side as the object).

Since the second lens $$L_2$$ is 10 cm to the right of $$L_1$$, this virtual image lies $$8 + 10 = 18$$ cm to the left of $$L_2$$ and thus acts as an object for $$L_2$$ with $$u_2 = -18$$ cm.

Applying the lens formula $$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$$ to $$L_2$$, we substitute $$u_2 = -18$$ cm and $$f_2 = 9$$ cm to obtain $$\dfrac{1}{v_2} - \dfrac{1}{-18} = \dfrac{1}{9}$$. This becomes $$\dfrac{1}{v_2} + \dfrac{1}{18} = \dfrac{1}{9}$$, so $$\dfrac{1}{v_2} = \dfrac{1}{9} - \dfrac{1}{18} = \dfrac{2-1}{18} = \dfrac{1}{18}$$, and hence $$v_2 = 18$$ cm. Consequently, the final image is 18 cm to the right of $$L_2$$.

Finally, since the object lies 6 cm to the left of $$L_1$$ and the final image lies 18 cm to the right of $$L_2$$, which itself is 10 cm to the right of $$L_1$$, the total separation between object and final image is $$6 + (10 + 18) = 6 + 28 = 34$$ cm. Thus the required distance is $$\boxed{34}$$ cm.

The angle of incidence $$i$$ with the normal is calculated from the angle made with the water surface:  $$i = 90^\circ - 30^\circ = 60^\circ$$

Applying Snell's law at the air-water interface to find the angle of refraction $$r$$:

$$\sin i = n_w \sin r$$

$$\sin 60^\circ = \frac{4}{3} \sin r \implies \frac{\sqrt{3}}{2} = \frac{4}{3} \sin r \implies \sin r = \frac{3\sqrt{3}}{8} \approx 0.65$$

$$\tan r = \frac{\sin r}{\sqrt{1 - \sin^2 r}} = \frac{\frac{3\sqrt{3}}{8}}{\sqrt{1 - \left(\frac{3\sqrt{3}}{8}\right)^2}} = \frac{3\sqrt{3}}{\sqrt{64 - 27}} = \frac{3\sqrt{3}}{\sqrt{37}} \approx 0.854$$

The total shadow length within the water is composed of the shadow component from the air section ($$x_1$$) and the water section ($$x_2$$):

$$x_{\text{total}} = x_1 + x_2$$ $$\implies$$ $$2.15 = h_a \tan 60^\circ + h_w \tan r$$

$$2.15 = h_a \sqrt{3} + 1.5 \times 0.854$$ $$\implies$$ $$2.15 = h_a \sqrt{3} + 1.281$$ $$\implies$$ $$h_a \sqrt{3} = 2.15 - 1.281 = 0.869$$

$$h_a = \frac{0.869}{\sqrt{3}} = \frac{0.869}{1.732} \approx 0.50\ \text{m}$$

$$h_a = 0.50 \times 100 = 50\ \text{cm}$$

Given: Glass plate thickness $$t = \sqrt{3}$$ cm, refractive index $$n = \sqrt{2}$$, angle of incidence = critical angle for glass-air interface.

The critical angle $$i_c$$ satisfies:

$$\sin i_c = \frac{1}{n} = \frac{1}{\sqrt{2}}$$

$$i_c = 45°$$

The angle of incidence $$i = 45°$$. Using Snell's law to find the angle of refraction inside the glass:

$$\sin i = n \sin r$$

$$\frac{1}{\sqrt{2}} = \sqrt{2} \sin r$$

$$\sin r = \frac{1}{2} \implies r = 30°$$

The lateral displacement formula for a parallel-sided glass slab is:

$$d = \frac{t \sin(i - r)}{\cos r}$$

Substituting the values:

$$d = \frac{\sqrt{3} \times \sin(45° - 30°)}{\cos 30°} = \frac{\sqrt{3} \times \sin 15°}{\cos 30°}$$

Using $$\sin 15° = 0.26$$ and $$\cos 30° = \frac{\sqrt{3}}{2}$$:

$$d = \frac{\sqrt{3} \times 0.26}{\frac{\sqrt{3}}{2}} = \frac{0.26 \times 2}{1} = 0.52 \text{ cm}$$

Converting to the required form:

$$d = 52 \times 10^{-2} \text{ cm}$$

Therefore, the answer is $$\mathbf{52}$$.

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies v = \frac{uf}{u - f}$$

Focal length of concave mirror $$f = -20 \text{ cm}$$

Length of rod $$L = 10 \text{ cm}$$

Midpoint of rod is at $$u_{\text{mid}} = -40 \text{ cm}$$

Since the midpoint is at $$40 \text{ cm}$$ from the pole, the rod extends $$5 \text{ cm}$$ on either side.

Position of the closer end ($$A$$): $$u_1 = -40 + 5 = -35 \text{ cm}$$

Position of the farther end ($$B$$): $$u_2 = -40 - 5 = -45 \text{ cm}$$

For end $$A$$ ($$u_1 = -35 \text{ cm}$$):

$$v_1 = \frac{(-35)(-20)}{-35 - (-20)}$$

$$v_1 = \frac{700}{-15}$$

$$v_1 = -\frac{140}{3} \text{ cm}$$

For end $$B$$ ($$u_2 = -45 \text{ cm}$$):

$$v_2 = \frac{(-45)(-20)}{-45 - (-20)}$$

$$v_2 = \frac{900}{-25}$$

$$v_2 = -36 \text{ cm}$$

The length of the image is the distance between these two points:

$$L' = |v_1 - v_2|$$

$$L' = \left| -\frac{140}{3} - (-36) \right|$$

$$L' = \left| -\frac{140}{3} + \frac{108}{3} \right|$$

$$L' = \left| -\frac{32}{3} \right| = \frac{32}{3} \text{ cm}$$

Lens maker formula used:

$$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

For the plano-concave lens,

$$\frac{1}{f_1}=(1.75-1)\left(-\frac{1}{30}\right)$$

$$\frac{1}{f_1}=0.75\left(-\frac{1}{30}\right)$$

$$f_1=-40\ \text{cm}$$

For the plano-convex lens,

$$\frac{1}{f_2}=(1.75-1)\left(\frac{1}{30}\right)$$

$$\frac{1}{f_2}=0.75\left(\frac{1}{30}\right)$$

$$f_2=40\ \text{cm}$$

The object is at infinity, so after refraction from the concave lens ($$L_1$$), the image is formed at its focal point.

Since $$f_1=-40$$ cm, the image formed by $$L_1$$ is virtual and $$40$$ cm to the left of $$L_1$$.

Distance between lenses:

$$L_1L_2 = 40\ \text{cm}$$

Therefore, object distance for $$L_2$$:

$$u_2 = 40+40=80\ \text{cm}$$

Since,

$$u_2=2f_2$$

Now we use the standard lens property:

When an object is placed at $$2f$$ of a convex lens, the image is formed at $$2f$$ on the other side.

So,

$$v_2=2f_2=80\ \text{cm}$$

You can also verify using lens formula:

$$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$

Putting,

$$f=40,\qquad u=-80$$

$$\frac{1}{40}=\frac{1}{v}+\frac{1}{80}$$

$$\frac{1}{v}=\frac{1}{40}-\frac{1}{80} = \frac{1}{80}$$

$$v=80\ \text{cm}$$

So image is formed at $$80\ \text{cm}$$ on the right side of the convex lens.

Hence total distance from concave lens:

x = 40 + 80

$$\boxed{x=120\ \text{cm}}$$

The power of a lens is given by the lensmaker's equation:

$$P = \left(\frac{n_l}{n_m} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

For a convex lens with both surfaces having radius of curvature 20 cm: $$R_1 = +20$$ cm, $$R_2 = -20$$ cm.

Power in air ($$n_m = 1$$):

$$P_{air} = (1.8 - 1)\left(\frac{1}{20} + \frac{1}{20}\right) = 0.8 \times \frac{2}{20} = 0.8 \times 0.1 = 0.08 \text{ cm}^{-1}$$

Power in liquid ($$n_m = 1.5$$):

$$P_{liq} = \left(\frac{1.8}{1.5} - 1\right)\left(\frac{1}{20} + \frac{1}{20}\right) = (1.2 - 1) \times 0.1 = 0.2 \times 0.1 = 0.02 \text{ cm}^{-1}$$

Ratio:

$$\frac{P_{air}}{P_{liq}} = \frac{0.08}{0.02} = 4$$

Therefore, $$x = 4$$.

For an equilateral prism ($$A = 60°$$), the refractive index and angle of minimum deviation are related by:

$$n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

$$\sqrt{2} = \frac{\sin\left(\frac{60° + D_m}{2}\right)}{\sin 30°} = \frac{\sin\left(\frac{60° + D_m}{2}\right)}{0.5}$$

$$\sin\left(\frac{60° + D_m}{2}\right) = \frac{\sqrt{2}}{2} = \sin 45°$$

$$\frac{60° + D_m}{2} = 45°$$

$$60° + D_m = 90°$$

$$D_m = 30°$$

The angle of minimum deviation is $$\mathbf{30}$$°.

$$\text{For the first lens: } u_1 = -\infty, \ f_1 = 20\text{ cm} \implies \frac{1}{v_1} - \frac{1}{-\infty} = \frac{1}{20} \implies v_1 = 20\text{ cm}$$

$$\text{The first image forms } 20\text{ cm to the right of the first lens, acting as the object for the second lens.}$$

$$\text{For the second lens: } u_2 = v_1 - d = 20 - 60 = -40\text{ cm}, \ f_2 = 20\text{ cm}$$

$$\frac{1}{v_2} - \frac{1}{u_2} = \frac{1}{f_2} \implies \frac{1}{v_2} - \frac{1}{-40} = \frac{1}{20} \implies \frac{1}{v_2} = \frac{1}{20} - \frac{1}{40} = \frac{1}{40} \implies v_2 = 40\text{ cm}$$

$$\text{Distance of the final image from the first lens: } D = d + v_2 = 60 + 40 = 100\text{ cm}$$

We have $$\mu_1 = 1.0$$ (rarer medium), $$\mu_2 = 1.5$$ (denser medium), radius of curvature $$R = 30$$ cm (centre of curvature towards the denser medium, so $$R = +30$$ cm), and object distance $$u = -15$$ cm (object in the rarer medium).

Using the refraction formula at a spherical surface:

$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$

Substituting the values:

$$\frac{1.5}{v} - \frac{1.0}{-15} = \frac{1.5 - 1.0}{30}$$

$$\frac{1.5}{v} + \frac{1}{15} = \frac{0.5}{30}$$

$$\frac{1.5}{v} + \frac{1}{15} = \frac{1}{60}$$

$$\frac{1.5}{v} = \frac{1}{60} - \frac{1}{15} = \frac{1}{60} - \frac{4}{60} = \frac{-3}{60} = \frac{-1}{20}$$

$$v = 1.5 \times (-20) = -30 \text{ cm}$$

The negative sign indicates the image is formed on the same side as the object (in the rarer medium), at a distance of 30 cm from the pole. So, the answer is $$30$$ cm.

$$\text{Apparent depth of the bulb when viewed from air above the water: } d_{\text{apparent}} = \frac{d_{\text{real}}}{\mu} = \frac{8}{4/3} = 6\text{ cm}$$

$$\text{Apparent shift upward: } \Delta x = d_{\text{real}} - d_{\text{apparent}} = 8 - 6 = 2\text{ cm}$$

$$\text{Effective distance from the bottom of the tank to the mirror: } H = 50\text{ cm}$$

$$\text{Effective object distance from the mirror: } u = H - \Delta x = 50 - 2 = 48\text{ cm}$$

$$\text{Since image distance equals object distance for a plane mirror, the image forms }48\text{ cm behind the mirror.}$$

$$\text{Distance of the image from the bottom of the tank: } D = H + v = 50 + 48 = 98\text{ cm}$$

$$\mu = \cot\left(\frac{A}{2}\right)$$

$$\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

$$\cot\left(\frac{A}{2}\right) = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$

$$\implies \frac{\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$$ $$\implies \cos\left(\frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right)$$

$$\sin\left(90^\circ - \frac{A}{2}\right) = \sin\left(\frac{A + \delta_m}{2}\right)$$ $$\implies 90^\circ - \frac{A}{2} = \frac{A + \delta_m}{2}$$

$$\implies 180^\circ - A = A + \delta_m$$ $$\implies \delta_m = 180^\circ - 2A$$

We need to find the angle of incidence for total internal reflection when light travels from a medium with dielectric constant $$\varepsilon_r = 4$$ to air.

The refractive index of the medium is given by $$n = \sqrt{\varepsilon_r \cdot \mu_r} = \sqrt{4 \times 1} = 2$$. At the critical angle $$\theta_c$$, the refracted ray travels along the surface (angle of refraction = $$90°$$), so $$\sin\theta_c = \frac{n_{air}}{n_{medium}} = \frac{1}{2}$$, which gives $$\theta_c = 30°$$.

Total internal reflection occurs when the angle of incidence is greater than or equal to the critical angle. For the entire incident wave to be reflected back, the angle of incidence must exceed $$30°$$. Among the given options ($$10°$$, $$20°$$, $$30°$$, $$60°$$), only $$60°$$ is greater than the critical angle, ensuring total internal reflection.

The correct answer is Option D: $$60°$$.

We are given that $$v_A - v_B = 2.6 \times 10^7$$ m/s, $$n_B = 1.47$$, and $$c = 3 \times 10^8$$ m/s. We need to find $$\frac{n_B}{n_A}$$.

First, the speed of light in medium B is $$v_B = \frac{c}{n_B} = \frac{3 \times 10^8}{1.47} = 2.0408 \times 10^8 \text{ m/s}$$. Since $$v_A - v_B = 2.6 \times 10^7$$ m/s, we have $$v_A = v_B + 2.6 \times 10^7 = 2.0408 \times 10^8 + 0.26 \times 10^8 = 2.3008 \times 10^8 \text{ m/s}$$.

The refractive index of medium A is then $$n_A = \frac{c}{v_A} = \frac{3 \times 10^8}{2.3008 \times 10^8} = 1.3039$$, and hence the ratio is $$\frac{n_B}{n_A} = \frac{1.47}{1.3039} = 1.1274 \approx 1.13$$. The correct answer is Option C: 1.13.

Light travels in medium $$M_1$$ with speed $$v_1 = 1.5 \times 10^8 \text{ m s}^{-1}$$ and in medium $$M_2$$ with speed $$v_2 = 2.0 \times 10^8 \text{ m s}^{-1}$$. To determine the critical angle between them, we begin by finding the refractive indices.

Since the speed of light in vacuum is $$c = 3 \times 10^8 \text{ m s}^{-1}$$, we have $$n_1 = \dfrac{c}{v_1} = \dfrac{3 \times 10^8}{1.5 \times 10^8} = 2$$ and $$n_2 = \dfrac{c}{v_2} = \dfrac{3 \times 10^8}{2.0 \times 10^8} = 1.5$$. Because $$n_1 > n_2$$, medium $$M_1$$ is denser than medium $$M_2$$ and total internal reflection occurs when light travels from $$M_1$$ to $$M_2$$.

At the critical angle $$\theta_c$$, the refracted ray emerges along the interface, so $$\sin \theta_c = \dfrac{n_2}{n_1} = \dfrac{1.5}{2} = \dfrac{3}{4}$$. This gives $$\sin \theta_c = \dfrac{3}{4}$$.

Next, we compute $$\cos \theta_c = \sqrt{1 - \sin^2 \theta_c} = \sqrt{1 - \dfrac{9}{16}} = \sqrt{\dfrac{7}{16}} = \dfrac{\sqrt{7}}{4}$$, which leads to $$\tan \theta_c = \dfrac{\sin \theta_c}{\cos \theta_c} = \dfrac{3/4}{\sqrt{7}/4} = \dfrac{3}{\sqrt{7}}$$. Therefore, $$\theta_c = \tan^{-1}\dfrac{3}{\sqrt{7}}$$.

Hence, the correct answer is Option A: $$\tan^{-1}\dfrac{3}{\sqrt{7}}$$.

We need to determine the refractive index of the surrounding medium ($$\mu_m$$) given the parameters of a biconvex lens operating inside that medium.

1. Identify the Given Parameters

• Power of the lens ($$P$$) = $$1.25\text{ m}^{-1}$$
• Refractive index of the lens material ($$\mu_l$$) = $$1.5$$
• Radius of curvature of the first surface ($$R_1$$) = $$20\text{ cm} = 0.2\text{ m}$$
• Radius of curvature of the second surface ($$R_2$$) = $$40\text{ cm} = 0.4\text{ m}$$

According to the standard Cartesian sign convention for a biconvex lens:

• The first refracting surface is convex, so its radius of curvature is positive: $$R_1 = +0.2\text{ m}$$.
• The second refracting surface is concave, so its radius of curvature is negative: $$R_2 = -0.4\text{ m}$$.

2. Apply the Lens Maker's Formula

The relationship between the power of a lens, its relative refractive index, and its radii of curvature is given by the Lens Maker's Formula:

$$P = \frac{1}{f} = \left(\frac{\mu_l}{\mu_m} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$

Substituting the known values into the formula:

$$1.25 = \left(\frac{1.5}{\mu_m} - 1\right) \left(\frac{1}{0.2} - \frac{1}{-0.4}\right)$$

3. Simplify and Solve for $$\mu_m$$

First, calculate the terms inside the geometric curvature bracket:

$$\left(\frac{1}{0.2} + \frac{1}{0.4}\right) = 5 + 2.5 = 7.5\text{ m}^{-1}$$

Substitute this value back into the main equation:

$$1.25 = \left(\frac{1.5}{\mu_m} - 1\right) \times 7.5$$

Isolate the bracket containing the refractive index variables:

$$\frac{1.25}{7.5} = \frac{1.5}{\mu_m} - 1$$

$$\frac{1}{6} = \frac{1.5}{\mu_m} - 1$$

Add 1 to both sides of the equation:

$$\frac{1.5}{\mu_m} = 1 + \frac{1}{6} = \frac{7}{6}$$

Now, rearrange to find $$\mu_m$$:

$$\mu_m = 1.5 \times \frac{6}{7} = \frac{3}{2} \times \frac{6}{7} = \frac{18}{14} = \frac{9}{7}$$

Conclusion

The refractive index of the surrounding medium is $$\frac{9}{7}$$. Based on the options provided in the problem layout, this directly matches Option B.

Correct Answer: Option B ($$\frac{9}{7}$$)

We need to identify the correct statement about a primary rainbow.

Option A: In primary rainbow, observer sees red colour on the top and violet on the bottom.

In a primary rainbow, sunlight enters a water droplet, undergoes one internal reflection, and exits. The red light is deviated less than violet light. Due to the geometry of the rainbow formation, the red arc appears on the outer (top) edge and the violet arc appears on the inner (bottom) edge. This is correct.

Option B: Violet on top and red on bottom.

This describes a secondary rainbow, not a primary one. This is incorrect.

Option C: Light suffers total internal reflection twice.

In a primary rainbow, light undergoes one internal reflection (not total internal reflection). Total internal reflection would not allow light to exit the drop. This is incorrect.

Option D: Primary rainbow is less bright than secondary rainbow.

The primary rainbow is brighter than the secondary rainbow because the secondary involves an additional reflection, losing more light intensity. This is incorrect.

Hence, the correct answer is Option A.

Concept:
Power of a lens depends on curvature, not on aperture (size).

Step 1: Lens cut along principal axis

• Focal length remains same ⇒ power unchanged

$$P_1 = P$$

Step 2: One half cut perpendicular to principal axis

• Radius of curvature effectively doubles ⇒ focal length doubles

$$P = \frac{1}{f} \Rightarrow P_{\text{new}} = \frac{1}{2f} = \frac{P}{2}$$​

So,

$$P_2 = \frac{P}{2}, \quad P_3 = \frac{P}{2}$$

Final Result:

$${P_1 = P,\quad P_2 = \frac{P}{2},\quad P_3 = \frac{P}{2}}$$

Given: Refractive index $$\mu = \sqrt{2n}$$, and the angle of incidence $$i = 2r$$ where $$r$$ is the angle of refraction.

Using Snell's law:

$$\sin i = \mu \sin r$$

$$\sin 2r = \sqrt{2n} \sin r$$

Using the identity $$\sin 2r = 2\sin r \cos r$$:

$$2\sin r \cos r = \sqrt{2n} \sin r$$

Since $$\sin r \neq 0$$, dividing both sides by $$\sin r$$:

$$2\cos r = \sqrt{2n}$$

$$\cos r = \frac{\sqrt{2n}}{2} = \sqrt{\frac{2n}{4}} = \sqrt{\frac{n}{2}}$$

$$r = \cos^{-1}\left(\sqrt{\frac{n}{2}}\right)$$

Since $$i = 2r$$:

$$i = 2\cos^{-1}\left(\sqrt{\frac{n}{2}}\right)$$

The correct answer is Option D.

An object at $$u = -2.4 \text{ m} = -240 \text{ cm}$$ forms a sharp image at $$v = 12 \text{ cm}$$. A glass plate of refractive index $$\mu = 1.5$$ and thickness $$t = 1 \text{ cm}$$ is introduced between the lens and screen.

Find the focal length of the lens.

Using the lens formula (with sign convention $$u = -240 \text{ cm}$$, $$v = +12 \text{ cm}$$):

Find the apparent shift caused by the glass plate.

The glass plate shifts the image towards the lens by:

Find the new image distance required.

For the image to still fall on the screen, the lens must form the image at:

Find the new object distance.

Calculate the shift in object position.

The object must be shifted from $$2.4 \text{ m}$$ to $$5.6 \text{ m}$$ away from the lens:

The correct answer is Option B: $$3.2 \text{ m}$$.

We have light entering from air into a medium at an angle of 45° with the interface. The angle of incidence is measured from the normal to the interface, so the angle of incidence is $$i = 90° - 45° = 45°$$.

After refraction, the light is deviated by 15° from its original direction. The deviation equals the difference between the angle of incidence and the angle of refraction: $$\delta = i - r$$. Therefore $$r = i - \delta = 45° - 15° = 30°$$.

Now applying Snell's law: $$n_1 \sin i = n_2 \sin r$$, where $$n_1 = 1$$ (air). So: $$\sin 45° = n_2 \sin 30°$$ $$\frac{\sqrt{2}}{2} = n_2 \cdot \frac{1}{2}$$ $$n_2 = \sqrt{2} \approx 1.414$$

Hence, the correct answer is Option 3.

Two materials A and B have refractive indices $$\mu_A$$ and $$\mu_B$$ with $$\mu_A : \mu_B = 1 : 2$$ and both have the same thickness $$d$$. The time difference between light traversing them is given as $$t_2 - t_1 = 5 \times 10^{-10} \text{ s}$$, and the individual times can be expressed as $$t_1 = \frac{d}{v_A}$$ and $$t_2 = \frac{d}{v_B}$$.

The refractive index is related to the speed of light in a medium by $$\mu = \frac{c}{v}$$, so that $$v = \frac{c}{\mu}$$. Given $$\mu_A : \mu_B = 1 : 2$$, it follows that $$\mu_B = 2\mu_A$$, hence $$v_A = \frac{c}{\mu_A}$$ and $$v_B = \frac{c}{\mu_B} = \frac{c}{2\mu_A}$$, which implies $$v_B = \frac{v_A}{2}$$.

Substituting these velocities into the expression for the time difference yields $$t_2 - t_1 = \frac{d}{v_B} - \frac{d}{v_A} = d\left(\frac{1}{v_B} - \frac{1}{v_A}\right) = d\left(\frac{2}{v_A} - \frac{1}{v_A}\right) = \frac{d}{v_A}\,.$$

Equating this to the given time difference, $$\frac{d}{v_A} = 5 \times 10^{-10}$$, gives $$d = 5 \times 10^{-10} \times v_A \text{ m}\,.$$

Hence, the correct answer is Option A.

Using Snell's law at successive parallel interfaces starting from air ($$\mu_0 = 1$$) at angle $$\theta = 60^\circ$$:

$$\mu_0 \sin\theta = \mu_1 \sin r_1 = \mu_2 \sin r_2 = \mu_{\text{air}} \sin r_{\text{air}}$$

$$\tan r_1 = 1 \implies r_1 = 45^\circ$$

$$\sin r_2 = \frac{1 \cdot \sin 60^\circ}{\sqrt{3}} = \frac{1}{2} \implies r_2 = 30^\circ$$

$$\sin r_{\text{air}} = \sin 60^\circ \implies r_{\text{air}} = 60^\circ$$

$$\Delta x_{\text{unit}} = h \tan r_{\text{air}} + d \tan r_1 + d \tan r_2$$

$$\Delta x_{\text{unit}} = \frac{1}{3}\sqrt{3} + \left(\frac{\sqrt{3}-1}{2}\right)(1) + \left(\frac{\sqrt{3}-1}{2}\right)\left(\frac{1}{\sqrt{3}}\right)$$

$$\Delta x_{\text{unit}} = \frac{1}{\sqrt{3}} + \left(\frac{\sqrt{3}-1}{2}\right)\left(1 + \frac{1}{\sqrt{3}}\right) = \frac{1}{\sqrt{3}} + \left(\frac{\sqrt{3}-1}{2}\right)\left(\frac{\sqrt{3}+1}{\sqrt{3}}\right)$$

$$\Delta x_{\text{unit}} = \frac{1}{\sqrt{3}} + \frac{3-1}{2\sqrt{3}} = \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}}\text{ cm}$$

$$l = n \cdot \Delta x_{\text{unit}}$$

$$\frac{8}{\sqrt{3}} = n \cdot \frac{2}{\sqrt{3}} \implies n = 4$$

Let the pole of the mirror be the origin and the principal axis be the positive $$x$$-axis directed towards the right. Hence distances measured towards the left are negative.

The focal length of a concave mirror is negative, so
$$f = -10\ \text{cm}$$

At the instant of interest the separation between the object and the mirror is
$$u_{\text{mag}} = 30\ \text{cm}$$

With the sign convention this becomes
$$u = -30\ \text{cm}$$

Using the mirror formula
$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}\quad -(1)$$

Substituting $$f = -10\ \text{cm}$$ and $$u = -30\ \text{cm}$$ in (1):
$$\frac{1}{-10} = \frac{1}{-30} + \frac{1}{v}$$
$$\frac{1}{v} = -\frac{1}{10} + \frac{1}{30} = -\frac{2}{30} = -\frac{1}{15}$$
$$\Rightarrow\ v = -15\ \text{cm}$$

Thus the image is formed $$15\ \text{cm}$$ in front of the mirror (real image, left side).

Let
$$x_m(t)$$ be the mirror position, $$x_o(t)$$ the object position, $$x_i(t)$$ the image position (all in the laboratory frame).

Object speed (towards the mirror):
$$V_o = +15\ \text{cm s}^{-1}$$

Mirror speed (to be found): $$V_m$$ (taken positive towards the right).

Signed object distance from the mirror is
$$u = x_o - x_m\quad\Rightarrow\quad \frac{du}{dt} = V_o - V_m = 15 - V_m$$

Signed image distance from the mirror is $$v$$. Differentiate the mirror formula $$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$ with respect to time:

$$0 = -\frac{1}{u^{2}}\frac{du}{dt} - \frac{1}{v^{2}}\frac{dv}{dt}$$
$$\Rightarrow\ \frac{dv}{dt} = -\frac{v^{2}}{u^{2}}\frac{du}{dt}$$

At the given instant $$u = -30\ \text{cm}$$ and $$v = -15\ \text{cm}$$, so
$$\frac{v^{2}}{u^{2}} = \frac{(-15)^{2}}{(-30)^{2}} = \frac{225}{900} = \frac14$$
Hence
$$\frac{dv}{dt} = -\frac14\left(15 - V_m\right)$$

The image position in the laboratory frame is
$$x_i = x_m + v$$ (because $$v$$ is measured from the mirror towards the left).
Therefore
$$\frac{dx_i}{dt} = V_m + \frac{dv}{dt}$$

The question states that the image is instantaneously at rest in the laboratory frame, so
$$\frac{dx_i}{dt} = 0$$
$$\Rightarrow\ V_m + \frac{dv}{dt} = 0$$

Insert $$\frac{dv}{dt}$$ from above:
$$V_m - \frac14\left(15 - V_m\right) = 0$$
Multiply by 4:
$$4V_m - (15 - V_m) = 0$$
$$5V_m - 15 = 0$$
$$\boxed{V_m = 3\ \text{cm s}^{-1}}$$

The mirror must move rightwards with speed $$3\ \text{cm s}^{-1}$$. Its magnitude is therefore 3 cm s$$^{-1}$$, which satisfies the additional requirement that the image remains real.

A ray of light is incident at $$60°$$ on a glass slab of refractive index $$\sqrt{3}$$. The lateral shift is $$4\sqrt{3}$$ cm. We need to find the thickness of the glass slab.

Using Snell’s law, we have:

$$n_1 \sin i = n_2 \sin r$$

Substituting the given values yields:

$$1 \times \sin 60° = \sqrt{3} \times \sin r$$

which simplifies to:

$$\frac{\sqrt{3}}{2} = \sqrt{3} \sin r$$

and hence:

$$\sin r = \frac{1}{2}$$

giving:

$$r = 30°$$

The lateral shift $$d$$ for a parallel glass slab is given by:

$$d = \frac{t}{\cos r} \sin(i - r)$$

Substituting $$i = 60°$$, $$r = 30°$$, and $$d = 4\sqrt{3}$$ cm, we get:

$$4\sqrt{3} = \frac{t}{\cos 30°} \sin(60° - 30°)$$

Since $$\cos 30° = \frac{\sqrt{3}}{2}$$ and $$\sin 30° = \frac{1}{2}$$, this becomes:

$$4\sqrt{3} = \frac{t}{\frac{\sqrt{3}}{2}} \times \frac{1}{2}$$

or equivalently:

$$4\sqrt{3} = \frac{t}{\sqrt{3}}$$

Multiplying both sides by $$\sqrt{3}$$ gives:

$$t = 4\sqrt{3} \times \sqrt{3} = 4 \times 3 = 12 \text{ cm}$$

Therefore, the answer is $$12$$ cm.

We have two identical thin biconvex lenses, each with focal length $$f = 15$$ cm and refractive index $$n = 1.5$$, in contact with each other. The space between the lenses is filled with a liquid of refractive index $$n_l = 1.25$$. Using the Lensmaker’s equation for a biconvex lens, $$\frac{1}{f} = (n - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ and noting that for identical biconvex lenses with equal radii of curvature $$R_1 = R$$ and $$R_2 = -R$$, we get:

$$\frac{1}{15} = (1.5 - 1)\left(\frac{1}{R} + \frac{1}{R}\right) = 0.5 \times \frac{2}{R} = \frac{1}{R}$$

$$R = 15 \text{ cm}$$

The liquid filling the space acts as a concave (diverging) lens with surfaces matching the inner surfaces of the biconvex lenses, having radii $$R_1 = -R = -15$$ cm and $$R_2 = R = 15$$ cm. For this liquid lens:

$$\frac{1}{f_l} = (n_l - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (1.25 - 1)\left(\frac{1}{-15} - \frac{1}{15}\right)$$

$$\frac{1}{f_l} = 0.25 \times \left(-\frac{2}{15}\right) = -\frac{0.5}{15} = -\frac{1}{30}$$

$$f_l = -30 \text{ cm}$$

Since the two convex lenses and the liquid lens are in contact, the combined focal length is given by

$$\frac{1}{f_{\text{comb}}} = \frac{1}{f_1} + \frac{1}{f_l} + \frac{1}{f_2} = \frac{1}{15} + \left(-\frac{1}{30}\right) + \frac{1}{15}$$

$$\frac{1}{f_{\text{comb}}} = \frac{2}{15} - \frac{1}{30} = \frac{4}{30} - \frac{1}{30} = \frac{3}{30} = \frac{1}{10}$$

$$f_{\text{comb}} = 10 \text{ cm}$$

The answer is 10 cm.

We need to determine the focal length of the convex mirror when it is placed in combination with a convex lens such that the final image coincides with the point object itself.

1. Analyze the First Refraction (Through the Convex Lens)

Using the lens formula, we find where the convex lens forms the first image:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f_{\text{lens}}}$$

From the problem

• Focal length of the convex lens ($$f_{\text{lens}}$$) = $$20\text{ cm}$$
• Object distance ($$u$$) = $$-60\text{ cm}$$ (using standard Cartesian sign convention)

Substitute these values to solve for the image distance ($$v$$):

$$\frac{1}{v} - \frac{1}{-60} = \frac{1}{20}$$

$$\frac{1}{v} + \frac{1}{60} = \frac{1}{20}$$

$$\frac{1}{v} = \frac{1}{20} - \frac{1}{60} = \frac{3 - 1}{60} = \frac{2}{60} = \frac{1}{30}$$

$$v = 30\text{ cm}$$

This means the lens forms a real image $$30\text{ cm}$$ behind it.

2. Analyze the Reflection (By the Convex Mirror)

The problem states that the final image retraces its path and coincides with the object itself. For a light ray to retrace its path after hitting a spherical mirror, it must strike the mirror surface normally (perpendicularly).

This happens only if the rays heading from the lens are directed straight toward the center of curvature ($$C$$) of the convex mirror.

• The distance from the lens to the image position is $$30\text{ cm}$$.
• The distance between the lens and the mirror is given as $$10\text{ cm}$$.

Therefore, the distance from the mirror to its center of curvature ($$R$$) is:

$$R = 30\text{ cm} - 10\text{ cm} = 20\text{ cm}$$

3. Calculate the Focal Length of the Mirror ($$f_{\text{mirror}}$$)

The relationship between the focal length and the radius of curvature for a spherical mirror is:

$$f_{\text{mirror}} = \frac{R}{2}$$

$$f_{\text{mirror}} = \frac{20}{2} = 10\text{ cm}$$

Conclusion

The focal length of the convex mirror is 10 cm.

We need to find the area of the water surface through which light from a bulb at the bottom can emerge.

Given that the depth of water is $$h = \sqrt{7}$$ m and the refractive index of water is $$\mu = \frac{4}{3}$$.

Since light can emerge from water only within the cone defined by the critical angle $$\theta_c$$, we have $$\sin \theta_c = \frac{1}{\mu} = \frac{3}{4}$$.

Now to determine the radius $$r$$ of the circle of light at the surface, note that $$\tan \theta_c = \frac{r}{h}$$. Using $$\sin \theta_c = \frac{3}{4}$$, we find $$\cos \theta_c = \sqrt{1 - \sin^2 \theta_c} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$, which gives $$\tan \theta_c = \frac{\sin \theta_c}{\cos \theta_c} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}}$$. Substituting into $$r = h \times \tan \theta_c$$ yields $$r = \sqrt{7} \times \frac{3}{\sqrt{7}} = 3$$ m.

Next, the area is given by $$A = \pi r^2 = \pi \times 3^2 = 9\pi$$ m$$^2$$. Comparing this with $$x\pi$$, we obtain $$x = 9$$.

The value of $$x$$ is 9.

We need to determine the value of $$x$$ given that the refracted light beam just grazes along the face $$AC$$ of an equilateral glass prism immersed in a liquid of refractive index $$n = \frac{\sqrt{x}}{4}$$.

1. Identify the Angles at Face $$AB$$ and Face $$AC$$

From the diagram 

• The prism is equilateral, so the angle of the prism is $$A = 60^\circ$$.
• The light ray enters normally at the first surface $$AB$$ because it enters perpendicular to that surface. Thus, the angle of refraction at the first surface is $$r_1 = 0^\circ$$.
• The relationship between the prism angle and the internal angles is given by $$A = r_1 + r_2$$. Substituting our values gives the angle of incidence at the second face $$AC$$:

  $$r_2 = A - r_1 = 60^\circ - 0^\circ = 60^\circ$$

• At the face $$AC$$, the refracted light beam just grazes along the surface, which means the angle of refraction in the liquid is $$r = 90^\circ$$. This implies $$60^\circ$$ is the critical angle for the glass-liquid interface.

2. Apply Snell's Law at Face $$AC$$ and Solve for $$x$$

Applying Snell's Law at the interface between the glass ($\mu_{\text{glass}} = 1.5 = \frac{3}{2}$) and the surrounding liquid ($$n$$):

$$\mu_{\text{glass}} \cdot \sin(r_2) = n \cdot \sin(r)$$

Substitute the known angles and values into the equation:

$$\frac{3}{2} \cdot \sin(60^\circ) = n \cdot \sin(90^\circ)$$

$$\frac{3}{2} \cdot \frac{\sqrt{3}}{2} = n \cdot 1 \implies n = \frac{3\sqrt{3}}{4}$$

Now, equate this result to the expression provided in the problem statement ($$n = \frac{\sqrt{x}}{4}$$):

$$\frac{\sqrt{x}}{4} = \frac{3\sqrt{3}}{4}$$

$$\sqrt{x} = 3\sqrt{3}$$

Squaring both sides of the equation to isolate $$x$$:

$$x = (3\sqrt{3})^2 = 9 \times 3 = 27$$

Conclusion

The value of $$x$$ is 27.

We need to find the value of the radius of curvature $$R$$ for a thin equi-convex lens based on the given graph between $$\frac{1}{v}$$ and $$\frac{1}{u}$$.

1. Relate the Graph to the Lens Formula

The standard lens formula for a thin lens relates the object distance ($$u$$), image distance ($$v$$), and focal length ($$f$$):

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

Rearranging this equation to express it in the slope-intercept straight-line form ($$y = mx + c$$), where $$\frac{1}{v}$$ is plotted on the vertical $$y$$-axis and $$\frac{1}{u}$$ is plotted on the horizontal $$x$$-axis:

$$\frac{1}{v} = \frac{1}{u} + \frac{1}{f}$$

From this equation:

• When the line intersects the vertical axis ($$\frac{1}{u} = 0$$ at point $$A$$), the $$y$$-intercept is equal to $$\frac{1}{f}$$.
• When the line intersects the horizontal axis ($$\frac{1}{v} = 0$$ at point $$B$$), the $$x$$-intercept is equal to $$-\frac{1}{f}$$.

2. Extract the Focal Length ($$f$$) from the Graph

Looking at the graph provided :

• The vertical intercept at point $$A$$ has a value of:

  $$\frac{1}{v} = 0.10\text{ cm}^{-1}$$

Equating this intercept to our formula gives:

$$\frac{1}{f} = 0.10\text{ cm}^{-1} \implies f = \frac{1}{0.10} = 10\text{ cm}$$

3. Calculate the Radius of Curvature ($$R$$)

According to the Lens Maker's Formula for a thin lens:

$$\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$

For an equi-convex lens under standard Cartesian sign conventions:

• The radius of curvature of the first surface is positive ($$R_1 = +R$$).
• The radius of curvature of the second surface is negative ($$R_2 = -R$$).

Substituting these along with the given refractive index ($$n = 1.5$$):

$$\frac{1}{f} = (1.5 - 1) \left( \frac{1}{R} - \left(-\frac{1}{R}\right) \right)$$

$$\frac{1}{f} = 0.5 \left( \frac{2}{R} \right) = \frac{1}{R}$$

Since $$\frac{1}{f} = \frac{1}{R}$$, it directly simplifies to:

$$R = f = 10\text{ cm}$$

Conclusion

The value of the radius of curvature $$R$$ is 10 cm.

We need to determine the position of the image formed by a concave mirror after an object 'O' has been moving towards it for 10 seconds.

Step 1: Identify the Initial Mirror Parameters

According to the standard Cartesian sign convention for a concave mirror:

• Radius of curvature ($$R$$) = $$-200\text{ cm}$$
• Focal length ($$f$$) = $$\frac{R}{2} = \frac{-200}{2} = -100\text{ cm}$$
• Initial object distance ($$u_{\text{initial}}$$) = $$-100\text{ cm}$$

Note: Initially, the object is placed exactly at the focal point of the mirror ($$u = f$$).

Step 2: Calculate the New Object Position after 10 Seconds

The object moves towards the mirror at a constant speed of $$2\text{ cm s}^{-1}$$.

• Distance traveled by the object in $10\text{ s}$ ($\Delta s$) = $$\text{speed} \times \text{time} = 2\text{ cm s}^{-1} \times 10\text{ s} = 20\text{ cm}$$

Since the object moves closer to the mirror, its new distance from the pole decreases in magnitude:

$$\text{New distance} = 100\text{ cm} - 20\text{ cm} = 80\text{ cm}$$

Applying the sign convention, the new object distance ($$u$$) is:

$$u = -80\text{ cm}$$

Step 3: Calculate the Image Position ($$v$$) Using the Mirror Formula

The relationship between the focal length, object distance, and image distance is given by the mirror formula:

$$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$$

Rearranging the formula to isolate $$\frac{1}{v}$$:

$$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$$

Substituting the values ($$f = -100\text{ cm}$$ and $$u = -80\text{ cm}$$):

$$\frac{1}{v} = \frac{1}{-100} - \frac{1}{-80}$$

$$\frac{1}{v} = -\frac{1}{100} + \frac{1}{80}$$

Finding a common denominator (400) to simplify the fractions:

$$\frac{1}{v} = \frac{-4 + 5}{400} = \frac{1}{400}$$

$$v = 400\text{ cm}$$

Conclusion

The positive sign indicates that a virtual image is formed behind the reflecting surface of the mirror.

Therefore, the position of the image from the mirror after 10 seconds will be at 400 cm.

We have the X-Y plane as the boundary between media $$M_1$$ ($$Z \geq 0$$, refractive index $$n_1 = \sqrt{2}$$) and $$M_2$$ ($$Z < 0$$, refractive index $$n_2 = \sqrt{3}$$). A ray travels in $$M_1$$ along the direction $$\vec{A} = 4\sqrt{3}\hat{i} - 3\sqrt{3}\hat{j} - 5\hat{k}$$.

The boundary is the X-Y plane, so the normal to the boundary is along the $$\hat{k}$$ direction. The angle of incidence is the angle between the ray and the normal to the surface. Since the ray has a negative $$k$$-component, it is moving toward the boundary (into the $$-z$$ direction).

The magnitude of $$\vec{A}$$ is $$|\vec{A}| = \sqrt{(4\sqrt{3})^2 + (-3\sqrt{3})^2 + (-5)^2} = \sqrt{48 + 27 + 25} = \sqrt{100} = 10$$.

The component of the ray along the normal ($$-\hat{k}$$) direction has magnitude 5, and the component along the surface (in the X-Y plane) has magnitude $$\sqrt{48 + 27} = \sqrt{75} = 5\sqrt{3}$$.

The angle of incidence $$i$$ satisfies $$\cos i = \dfrac{|\text{component along normal}|}{|\vec{A}|} = \dfrac{5}{10} = \dfrac{1}{2}$$, so $$i = 60°$$.

Now applying Snell's law: $$n_1 \sin i = n_2 \sin r$$, so $$\sqrt{2} \sin 60° = \sqrt{3} \sin r$$. We get $$\sqrt{2} \times \dfrac{\sqrt{3}}{2} = \sqrt{3} \sin r$$, which simplifies to $$\dfrac{\sqrt{6}}{2} = \sqrt{3} \sin r$$, giving $$\sin r = \dfrac{\sqrt{6}}{2\sqrt{3}} = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}$$. Therefore $$r = 45°$$.

The difference between the angle of incidence and the angle of refraction is $$i - r = 60° - 45° = 15°$$.

Hence, the correct answer is 15.

We need to determine the value of $$x$$ given that the net average deviation produced by a combination of two thin prisms is equal to $$\left(\frac{1}{x}\right)^\circ$$.

1. Identify the Given Parameters

• Angle of the first prism ($$A_1$$) = $$6^\circ$$
• Refractive index of the first prism for yellow light ($$n_1$$) = $$1.5$$
• Angle of the second prism ($$A_2$$) = $$5^\circ$$
• Refractive index of the second prism for yellow light ($$n_2$$) = $$1.55$$

2. Understand Deviation Through Thin Prisms

The average (or net) deviation ($\delta$) produced by a single thin prism for mean (yellow) light is given by the formula:

$$\delta = (n - 1)A$$

In this non-dispersive setup, the two prisms are placed inverted relative to each other (as shown in the diagram), meaning their individual deviations oppose one another. The net average deviation ($$\delta_{\text{net}}$$) is the difference between their individual deviations:

$$\delta_{\text{net}} = \delta_1 - \delta_2$$

3. Calculate Individual Deviations

First, let's find the deviation produced by the first prism ($$\delta_1$$):

$$\delta_1 = (1.5 - 1) \times 6^\circ = 0.5 \times 6^\circ = 3^\circ$$

Next, let's find the deviation produced by the second prism ($$\delta_2$$):

$$\delta_2 = (1.55 - 1) \times 5^\circ = 0.55 \times 5^\circ = 2.75^\circ$$

4. Calculate Net Average Deviation and Solve for $$x$$

Subtracting the two values gives the total net deviation of the combination:

$$\delta_{\text{net}} = 3^\circ - 2.75^\circ = 0.25^\circ$$

We can convert this decimal value into a fractional form to match the format given in the problem statement:

$$\delta_{\text{net}} = \left(\frac{25}{100}\right)^\circ = \left(\frac{1}{4}\right)^\circ$$

Comparing this result with the expression $$\left(\frac{1}{x}\right)^\circ$$ 

$$\frac{1}{x} = \frac{1}{4} \implies x = 4$$

Conclusion

The value of $$x$$ is 4.

We are given that when the image distance $$v'$$ and the object distance $$u'$$ are measured from the focus of a convex lens, a plot of $$v'$$ against $$u'$$ gives the curve $$v' u' = 225$$, with all distances in cm. We need to find the focal length of the lens.

This problem uses Newton's lens formula. To derive it, recall the standard thin lens equation $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$, where $$v$$ is the image distance and $$u$$ is the object distance, both measured from the optical centre of the lens.

Now, let us define $$u'$$ and $$v'$$ as the distances measured from the respective foci. If the object is at distance $$u'$$ from the first focus (on the object side), then $$u = -(f + u')$$ (taking the sign convention where the object is on the negative side). Similarly, if the image is at distance $$v'$$ from the second focus (on the image side), then $$v = f + v'$$.

Substituting into the lens equation: $$\frac{1}{f + v'} + \frac{1}{f + u'} = \frac{1}{f}$$.

Taking the LCM on the left side: $$\frac{(f + u') + (f + v')}{(f + v')(f + u')} = \frac{1}{f}$$.

Cross-multiplying: $$f(2f + u' + v') = (f + v')(f + u') = f^2 + f u' + f v' + v' u'$$.

Expanding the left side: $$2f^2 + f u' + f v' = f^2 + f u' + f v' + v' u'$$.

Cancelling the common terms $$f u'$$ and $$f v'$$ from both sides: $$2f^2 = f^2 + v' u'$$, which gives us $$v' u' = f^2$$.

This is Newton's formula for a thin lens: the product of the object and image distances measured from the respective foci equals the square of the focal length.

Given $$v' u' = 225$$, we have $$f^2 = 225$$. Taking the square root, $$f = \sqrt{225} = 15 \text{ cm}$$.

Hence, the magnitude of the focal length of the lens is $$\textbf{15}$$ cm.

(A) Ray will come out after one reflection for $$\theta = 30^\circ$$ & $$0 < \ell < L$$.

(B) For $$\theta = 60^\circ$$ & $$\ell = \frac{L}{2}$$, ray will come out after two reflections.

(C) For $$\ell = \frac{L}{3}$$ & $$\theta = 60^\circ$$, ray will come out after five reflections.

(D) For $$\theta = 60^\circ$$ & $$0 < \ell < \frac{L}{2}$$, ray will come out after five reflections.

For every combination of the two-lens system we shall use the Cartesian sign convention and the thin-lens formula
$$\frac1v - \frac1u = \frac1f \qquad\text{-(1)}$$
Here $$u$$ is the object distance from the lens, $$v$$ the image distance from the lens (both measured from the lens‐centre, positive to the right), and $$f$$ the focal length (positive for a converging lens, negative for a diverging lens).
The lenses are 5 cm apart and in each case the original object is 20 cm to the left of lens 1, so $$u_1 = -20\text{ cm}$$ for every combination.

Lens 1: $$u_1 = -20\text{ cm},\; f_1 = +20\text{ cm}$$
Using (1): $$\frac1{v_1} = \frac1{20} + \frac1{-20} = 0 \;\Rightarrow\; v_1 = +\infty$$
Rays emerging from lens 1 are parallel, so for lens 2 the object is at infinity: $$u_2 = -\infty$$.
Lens 2 therefore forms the image at its own focus:
$$v_2 = f_2 = +7.5\text{ cm}$$ (to the right of lens 2).
Hence the final image position matches item P.

Lens 1: $$u_1 = -20\text{ cm},\; f_1 = +10\text{ cm}$$
$$\frac1{v_1} = \frac1{10} + \frac1{-20} = \frac1{20}\;\Rightarrow\; v_1 = +20\text{ cm}$$
This image lies 20 cm to the right of lens 1. The two lenses are 5 cm apart, therefore the object distance for lens 2 is
$$u_2 = -(20 - 5) = -15\text{ cm}$$
Lens 2: $$f_2 = +30\text{ cm}$$
$$\frac1{v_2} = \frac1{30} + \frac1{-15} = \frac1{30} - \frac2{30} = -\frac1{30}$$
$$\therefore\; v_2 = -30\text{ cm}$$ (30 cm to the left of lens 2).
Thus the final image corresponds to item R.

Lens 1: $$u_1 = -20\text{ cm},\; f_1 = +20\text{ cm}$$
Again $$v_1 = +\infty$$, so the rays reaching lens 2 are parallel: $$u_2 = -\infty$$.
Lens 2 therefore brings them to focus at
$$v_2 = f_2 = +60\text{ cm}$$ (to the right of lens 2).
This matches item Q.

Lens 1: $$u_1 = -20\text{ cm},\; f_1 = +10\text{ cm}$$
$$\frac1{v_1} = \frac1{10} + \frac1{-20} = \frac1{20}\;\Rightarrow\; v_1 = +20\text{ cm}$$
Hence $$u_2 = -(20 - 5) = -15\text{ cm}$$ for lens 2.
Lens 2: $$f_2 = +10\text{ cm}$$
$$\frac1{v_2} = \frac1{10} + \frac1{-15} = \frac3{30} - \frac2{30} = \frac1{30}$$
$$\therefore\; v_2 = +30\text{ cm}$$ (to the right of lens 2).
Thus the final image position coincides with item T.

Summarising the results:
I → P (7.5 cm to the right of lens 2)
II → R (30 cm to the left of lens 2)
III → Q (60 cm to the right of lens 2)
IV → T (30 cm to the right of lens 2)

The only option giving this correspondence is
Option A: I → P, II → R, III → Q, IV → T.

We begin by listing the known quantities. The refractive index of diamond is $$n_1 = 2.42$$, the refractive index of air is $$n_2 = 1.00$$ and the angle of incidence (measured from the normal) is $$i = 30^\circ$$. A ray is travelling from the denser medium (diamond) to the rarer medium (air).

Whenever light passes from one medium to another, the relationship between the angle of incidence $$i$$ and the angle of refraction $$r$$ is given by Snell’s law, which states

$$n_1 \sin i \;=\; n_2 \sin r.$$

Substituting the given numerical values, we have

$$2.42 \,\sin 30^\circ \;=\; 1.00 \,\sin r.$$

Now, $$\sin 30^\circ = \dfrac{1}{2}$$, so the left-hand side becomes

$$2.42 \times \dfrac{1}{2} = 1.21.$$

Hence the equation reduces to

$$1.21 = \sin r.$$

But the sine of any real angle can never exceed 1, i.e. $$\sin r \le 1$$ for all real $$r$$. Since $$1.21 > 1$$, the condition $$\sin r = 1.21$$ is physically impossible. This tells us that the ray cannot refract into air; instead, total internal reflection must occur.

To reinforce this conclusion, we can calculate the critical angle $$\theta_c$$ for the diamond-air interface. Total internal reflection happens when the angle of incidence exceeds this critical angle. The critical angle is obtained by setting $$r = 90^\circ$$ in Snell’s law:

$$n_1 \sin \theta_c = n_2 \sin 90^\circ.$$

Since $$\sin 90^\circ = 1$$, we get

$$\sin \theta_c = \dfrac{n_2}{n_1} = \dfrac{1.00}{2.42} \approx 0.413.$$

Therefore

$$\theta_c = \sin^{-1}(0.413) \approx 24.4^\circ.$$

The given angle of incidence is $$30^\circ$$, which is clearly greater than $$24.4^\circ$$. Hence the ray is indeed incident at an angle larger than the critical angle, confirming total internal reflection and the impossibility of refraction.

Hence, the correct answer is Option C.

We wish to find the image distance $$v$$ and the linear magnification $$m$$ when an object is kept at the focus of a concave (diverging) lens whose focal length is $$f$$. Throughout we follow the Cartesian sign convention: distances measured to the left of the lens are taken negative, to the right positive; for a concave lens the focal length is negative.

The object is at the focus, so the magnitude of the object distance is $$f$$, but the object lies to the left of the lens. Therefore

$$u = -\,f.$$

For any thin lens we have the lens formula, which we first state:

$$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}.$$

Now we substitute $$u = -\,f$$ and remember that for a concave lens $$f$$ itself is negative. Substituting, we get

$$\frac{1}{v} - \frac{1}{(-\,f)} = \frac{1}{(-\,f)}.$$

The second term simplifies because a minus sign in the denominator changes sign:

$$\frac{1}{v} + \frac{1}{f} = -\,\frac{1}{f}.$$

We now collect like terms:

$$\frac{1}{v} = -\,\frac{1}{f} - \frac{1}{f}.$$

Adding the right-hand side, we obtain

$$\frac{1}{v} = -\,\frac{2}{f}.$$

Taking the reciprocal of both sides gives

$$v = -\,\frac{f}{2}.$$

The negative sign again tells us that the image is formed on the same side as the object (to the left of the lens). Hence the distance of the image from the optical centre is

$$|v| = \frac{f}{2}.$$

Next, we calculate the magnification. The formula for linear magnification produced by a lens is

$$m = \frac{v}{u}.$$

Substituting $$v = -\,f/2$$ and $$u = -\,f$$, we have

$$m = \frac{-\,f/2}{-\,f} = \frac{1}{2}.$$

Thus the image is virtual, upright (positive magnification), half the size of the object, and situated at a distance $$f/2$$ from the lens on the same side as the object.

Hence, the correct answer is Option B.

The focal length of a lens in a medium is given by the lensmaker's equation: $$\frac{1}{f} = \left(\frac{n_{\text{lens}}}{n_{\text{medium}}} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$.

When the lens with refractive index $$n_{\text{lens}} = 1.4$$ is placed in a medium of the same refractive index $$n_{\text{medium}} = 1.4$$, the factor becomes $$\frac{n_{\text{lens}}}{n_{\text{medium}}} - 1 = \frac{1.4}{1.4} - 1 = 1 - 1 = 0$$.

Therefore, $$\frac{1}{f} = 0 \times \left(\frac{1}{R_1} - \frac{1}{R_2}\right) = 0$$, which means $$f = \infty$$.

When a lens is immersed in a medium having the same refractive index as the lens material, it loses its converging or diverging ability and behaves like a flat piece of glass. The focal length becomes infinite.

For a plano-convex lens, one surface is flat (radius $$\infty$$) and the other is curved with radius $$R$$. The diameter is 6 cm, so the radius of the lens aperture is $$r = 3$$ cm = 0.03 m. The thickness at the centre is $$t = 3$$ mm = 0.003 m.

Using the geometry of the curved surface, for a spherical surface of radius $$R$$ with a chord of half-length $$r$$ and sagitta (depth) $$t$$, we have $$R = \frac{r^2}{2t} + \frac{t}{2}$$. Since $$t \ll r$$, we approximate $$R \approx \frac{r^2}{2t} = \frac{(0.03)^2}{2 \times 0.003} = \frac{9 \times 10^{-4}}{6 \times 10^{-3}} = 0.15$$ m.

The refractive index of the lens material is $$\mu = \frac{c}{v} = \frac{3 \times 10^8}{2 \times 10^8} = 1.5$$. Using the lensmaker's equation for a plano-convex lens: $$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) = (1.5 - 1)\left(\frac{1}{0.15} - 0\right) = 0.5 \times \frac{1}{0.15} = \frac{0.5}{0.15} = \frac{10}{3}$$.

Therefore $$f = \frac{3}{10} = 0.3$$ m = 30 cm. The correct answer is option 4: 30 cm.

A ray travels from denser medium to rarer medium at angle of incidence $$i$$. The angle of reflection equals the angle of incidence, so $$r = i$$.

The reflected and refracted rays are perpendicular to each other, so: $$r + r' = 90°$$ $$r' = 90° - r = 90° - i$$

By Snell's law at the interface (denser to rarer): $$n_1 \sin i = n_2 \sin r'$$

where $$n_1$$ is the refractive index of the denser medium and $$n_2$$ is that of the rarer medium. Assuming the rarer medium is air ($$n_2 = 1$$): $$n_1 \sin i = \sin r' = \sin(90° - i) = \cos i$$ $$n_1 = \frac{\cos i}{\sin i} = \cot i = \cot r$$

The critical angle $$\theta_c$$ is defined by $$\sin\theta_c = \frac{n_2}{n_1} = \frac{1}{n_1}$$, so: $$\sin\theta_c = \frac{1}{\cot r} = \tan r$$ $$\theta_c = \sin^{-1}(\tan r)$$

In a simple microscope (magnifying glass), a convex lens of short focal length is used. The object is placed between the lens and its focus, and the lens produces a virtual, erect, and magnified image on the same side as the object.

Consider a thin convex lens. When a ray passes through the optical centre of the lens, it goes undeviated. This means that the object and its virtual image subtend the same angle at the optical centre of the lens. Since the eye is placed very close to the lens, the angular size of the object (as seen through the lens) is essentially equal to the angular size of the image. Therefore, Assertion A is true.

Now consider viewing the object without the lens. The closest distance at which the unaided eye can focus is the near point, $$D = 25$$ cm. Without the magnifying glass, the object must be placed at least 25 cm away, where it subtends a small angle $$\alpha_0 \approx \frac{h}{D}$$, with $$h$$ being the object size.

With the magnifying glass, the object can be placed much closer to the eye (between the lens and its focus, at a distance $$u < f$$, which is typically much less than 25 cm). At this closer distance, the object subtends a larger angle $$\alpha \approx \frac{h}{u}$$ at the lens (and hence at the eye). The magnifying power is $$m = \frac{\alpha}{\alpha_0} = \frac{D}{u}$$, which is greater than 1 since $$u < D$$. Therefore, Reason R is true.

Since the magnification arises precisely because the lens allows the object to be placed closer than the near point, thereby subtending a larger angle, Reason R correctly explains Assertion A. Both A and R are true and R is the correct explanation of A.

$$\mu_1 = 1.25,\ \mu_2 = 1.4,\ u = -40\ \text{cm},\ R = -25\ \text{cm}$$

$$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$

$$\frac{1.4}{v} - \frac{1.25}{-40} = \frac{1.4 - 1.25}{-25}$$ $$\implies \frac{1.4}{v} + \frac{1.25}{40} = \frac{0.15}{-25}$$

$$\implies \frac{1.4}{v} = -\frac{0.15}{25} - \frac{1.25}{40} = -\frac{3}{500} - \frac{1}{32}$$

$$\implies \frac{1.4}{v} = -\frac{24 + 125}{4000} = -\frac{149}{4000}$$ $$\implies v = -\frac{1.4 \times 4000}{149} = -\frac{5600}{149} \approx -37.58\ \text{cm}$$

$$|v| = 37.58\ \text{cm}$$

We need to determine the maximum value of the angle of incidence $$\theta$$ such that the light ray undergoes total internal reflection (TIR) at the adjacent vertical interface.

1. Understand the Path of the Light Ray

From the problem statement:

• The light ray enters from air ($$n_{\text{air}} = 1$$) into a denser rectangular slab with a refractive index $$n = \frac{4}{3}$$ at an angle of incidence $$\theta$$.
• Let $$r$$ be the angle of refraction inside the slab at this horizontal surface.
• The ray then travels through the block and hits the adjacent vertical surface. From basic geometry, the angle of incidence at this second vertical wall is $$90^\circ - r$$.

2. Apply the Condition for Total Internal Reflection

For total internal reflection to occur at the vertical wall boundary with air, the angle of incidence must be greater than or equal to the critical angle ($$\theta_c$$):

$$90^\circ - r \ge \theta_c \implies \sin(90^\circ - r) \ge \sin\theta_c$$

$$\cos r \ge \frac{1}{n}$$

To express this in terms of sine, we use the trigonometric identity $$\cos r = \sqrt{1 - \sin^2 r}$$:

$$\sqrt{1 - \sin^2 r} \ge \frac{1}{n} \implies 1 - \sin^2 r \ge \frac{1}{n^2} \implies \sin^2 r \le 1 - \frac{1}{n^2}$$

3. Relate to the Initial Boundary (Snell's Law)

Applying Snell's Law at the first horizontal boundary where light enters from air into the medium:

$$1 \cdot \sin\theta = n \cdot \sin r \implies \sin r = \frac{\sin\theta}{n}$$

Substitute this expression into our inequality condition from Step 2:

$$\left(\frac{\sin\theta}{n}\right)^2 \le 1 - \frac{1}{n^2} \implies \frac{\sin^2\theta}{n^2} \le \frac{n^2 - 1}{n^2}$$

Cancel out the common $$n^2$$ terms in the denominators:

$$\sin^2\theta \le n^2 - 1 \implies \sin\theta \le \sqrt{n^2 - 1}$$

4. Calculate the Numerical Value

Substitute the given refractive index $$n = \frac{4}{3}$$ into the formula to find the maximum angle constraint:

$$\sin\theta_{\text{max}} = \sqrt{\left(\frac{4}{3}\right)^2 - 1} = \sqrt{\frac{16}{9} - 1} = \sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3}$$

Isolating $$\theta$$ gives the maximum allowable limit:

$$\theta = \sin^{-1}\left(\frac{\sqrt{7}}{3}\right)$$

Conclusion

The maximum value of the angle $$\theta$$ must be equal to $$\sin^{-1}\left(\frac{\sqrt{7}}{3}\right)$$, which corresponds exactly to Option A.

We need to determine the correct conditions for a prism under which the angle of deviation $$\delta$$ is at its minimum value.

The angle of deviation $$\delta$$ through a triangular prism depends on the angle of incidence $$i$$ and follows a characteristic curve that reaches a single minimum point, known as the minimum deviation position $$\delta_m$$.

At the position of minimum deviation, the light ray passes through the prism symmetrically. This symmetry leads to several key physical conditions:

1. The angle of incidence $$i$$ becomes exactly equal to the angle of emergence $$e$$: $$i = e$$. This satisfies statement (C).

2. Because the angles are equal at both faces ($$i = e$$ and consequently the internal refraction angles $$r_1 = r_2$$), the incident ray and the emergent ray are completely symmetric with respect to the prism. This satisfies statement (A).

3. For a standard principal section of an isosceles or equilateral prism, this internal geometric symmetry forces the refracted ray inside the prism to travel parallel to the base of the prism. This satisfies statement (B).

4. There is no physical law or condition stating that the angle of emergence must be double the angle of incidence ($$e = 2i$$) at minimum deviation; in fact, they must be equal ($$e = i$$). Thus, statement (D) is incorrect.

Since statements (A), (B), and (C) are true, the correct combination matching the criteria is represented by Option A.

Therefore, the correct answer is Option A: Statements (A), (B) and (C) are true.

We need to find the wavelength ($$\lambda$$) for which light rays incident on the interface $$BC$$ pass through without bending (no deviation).

1. Condition for No Bending at the Interface

For a light ray to pass through an interface between two optical mediums without shifting or bending at any angle of incidence, the refractive indices of both mediums must be perfectly equal:

$$n_1 = n_2$$

2. Equate the Refractive Index Expressions

From the problem details , the equations for $$n_1$$ and $$n_2$$ are given as:

$$1.2 + \frac{10.8 \times 10^{-14}}{\lambda^2} = 1.45 + \frac{1.8 \times 10^{-14}}{\lambda^2}$$

Rearrange the terms to group the constant values on one side and the $\lambda$ terms on the other:

$$\frac{10.8 \times 10^{-14}}{\lambda^2} - \frac{1.8 \times 10^{-14}}{\lambda^2} = 1.45 - 1.2$$

$$\frac{(10.8 - 1.8) \times 10^{-14}}{\lambda^2} = 0.25$$

$$\frac{9 \times 10^{-14}}{\lambda^2} = 0.25$$

3. Solve for Wavelength ($$\lambda$$)

Isolate $$\lambda^2$$ in the equation:

$$\lambda^2 = \frac{9 \times 10^{-14}}{0.25}$$

$$\lambda^2 = 36 \times 10^{-14}\text{ m}^2$$

Taking the square root of both sides gives the value in meters:

$$\lambda = \sqrt{36 \times 10^{-14}} = 6 \times 10^{-7}\text{ m}$$

Convert the calculated metric value into nanometers ($1\text{ nm} = 10^{-9}\text{ m}$):

$$\lambda = 600 \times 10^{-9}\text{ m} = 600\text{ nm}$$

Conclusion

The wavelength for which the rays pass through the interface without bending is 600 nm.

```

Distance of virtual image $$I$$ behind the mirror: $$v = u = 60\ \text{cm}$$

Total distance from virtual image to the observer's line: $$x = v + D = 60 + 120 = 180\ \text{cm}$$

Using similar triangles properties: $$\frac{L}{d} = \frac{x}{v}$$

$$\implies \frac{L}{50} = \frac{180}{60}$$

$$\implies L = 50 \times 3 = 150\ \text{cm}$$

We need to find the total distance between the object and its final real image when the convex mirror is removed from the system.

1. Analyze the First Case (With the Convex Mirror)

We are given that the final image of the object coincides with the object itself. This means light rays traveling from the object pass through the convex lens, strike the convex mirror, and trace their paths exactly back to the object.

For light rays to retrace their path after reflecting off a convex mirror, they must hit the mirror surface normally (perpendicularly). This happens only if the refracted rays from the lens are traveling toward the center of curvature ($C$) of the convex mirror.

• Focal length of the convex mirror ($f_m$): $$15\text{ cm}$$
• Radius of curvature of the mirror ($R$):

  $$R = 2 \times f_m = 2 \times 15\text{ cm} = 30\text{ cm}$$

Therefore, in the absence of the mirror, the lens would form a virtual image at the center of curvature of the mirror. Let's calculate this position ($v$) relative to the lens:

$$\text{Distance from lens to mirror} = 8\text{ cm}$$

$$\text{Distance from mirror to } C = 30\text{ cm}$$

$$v = 8\text{ cm} + 30\text{ cm} = 38\text{ cm}$$

So, the image distance for the lens is $$v = +38\text{ cm}$$.

2. Analyze the Second Case (Mirror is Removed)

When the convex mirror is removed, the light rays simply continue along their path unobstructed and converge to form a real image at that very same position ($v = 38\text{ cm}$) from the lens.

Let's map out the positions of the key elements along the principal axis:

• The object is placed at a distance of $$12\text{ cm}$$ to the left of the lens.
• The real image is formed at a distance of $$38\text{ cm}$$ to the right of the lens.

3. Calculate the Distance Between Object and Image

The total horizontal distance ($D$) from the object to its final real image is the sum of the object distance and the image distance relative to the lens:

$$D = \text{Distance from Object to Lens} + \text{Distance from Lens to Image}$$

$$D = 12\text{ cm} + 38\text{ cm} = 50\text{ cm}$$

Final Answer: 50

At minimum deviation through a prism of angle $$A$$, the refracted ray inside the prism travels parallel to the base, and the angle of refraction inside the prism equals $$A/2$$. The angle of incidence at the first surface equals the angle of emergence at the second surface.

The problem states that the angle of incidence $$i$$ is double the angle of refraction $$r$$ inside the prism, i.e. $$i = 2r$$. Since at minimum deviation $$r = A/2$$, we have:

$$i = 2 \times \frac{A}{2} = A$$

Applying Snell's law at the first surface (air to prism):

$$\sin i = \mu \sin r$$

$$\sin A = \sqrt{3}\,\sin\!\left(\frac{A}{2}\right)$$

Using the identity $$\sin A = 2\sin(A/2)\cos(A/2)$$:

$$2\sin\!\left(\frac{A}{2}\right)\cos\!\left(\frac{A}{2}\right) = \sqrt{3}\,\sin\!\left(\frac{A}{2}\right)$$

Dividing both sides by $$\sin(A/2)$$ (non-zero):

$$2\cos\!\left(\frac{A}{2}\right) = \sqrt{3} \implies \cos\!\left(\frac{A}{2}\right) = \frac{\sqrt{3}}{2}$$

$$\frac{A}{2} = 30^\circ \implies A = 60^\circ$$

The angle of the prism is $$60^\circ$$.

In the first scenario, an object is viewed at the near point distance $$D = 25$$ cm using a single microscopic lens, giving a magnification of $$m_0 = 6$$. For a simple lens with the image at the near point, the magnification is $$m_0 = 1 + \frac{D}{f_{obj}}$$, giving $$6 = 1 + \frac{25}{f_{obj}}$$, so $$f_{obj} = 5$$ cm.

In the second scenario, a compound microscope is used with an eyepiece and a tube of length $$L = 0.6$$ m $$= 60$$ cm. The image is now observed at infinity (relaxed eye), and the total magnification is double the earlier, i.e., $$m_{total} = 12$$.

For a compound microscope with final image at infinity, the total magnification is:

where $$L$$ is the tube length (length of the microscope tube, taken as the image distance from the objective minus its focal length). Substituting $$f_{obj} = 5$$ cm, $$D = 25$$ cm, $$L = 60$$ cm, and $$m_{total} = 12$$:

Therefore, the focal length of the eyepiece is $$\boxed{25}$$ cm.

The wavelength inside the surface is $$\frac{2}{3}$$ times the wavelength in air, which means the refractive index of the medium is $$\mu = \frac{\lambda_{\text{air}}}{\lambda_{\text{medium}}} = \frac{3}{2}$$.

The image is real and located 10 m behind the surface (in the denser medium), so $$v = +10$$ m. The image distance is $$\frac{2}{3}$$ of the object distance, meaning $$v = \frac{2}{3}|u|$$, so $$|u| = \frac{3}{2} \times 10 = 15$$ m and $$u = -15$$ m (object is on the same side as the incoming light).

Using the refraction formula for a single spherical surface, $$\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$$, where $$\mu_1 = 1$$ (air), $$\mu_2 = \frac{3}{2}$$, $$v = 10$$, $$u = -15$$, and $$R$$ is positive for a convex surface.

Substituting: $$\frac{3/2}{10} - \frac{1}{-15} = \frac{3/2 - 1}{R}$$, which gives $$\frac{3}{20} + \frac{1}{15} = \frac{1}{2R}$$. Computing the left side: $$\frac{3}{20} + \frac{1}{15} = \frac{9}{60} + \frac{4}{60} = \frac{13}{60}$$.

So $$\frac{13}{60} = \frac{1}{2R}$$, giving $$R = \frac{60}{26} = \frac{30}{13}$$ m. Since $$R = \frac{x}{13}$$, we get $$x = 30$$.

We have an achromatic combination of a crown glass prism and a flint glass prism that produces a net deviation of $$\delta = 2°$$ for yellow light. The given data is: dispersive power of crown glass $$\omega_c = 0.02$$, dispersive power of flint glass $$\omega_f = 0.03$$, refractive index of crown glass for yellow light $$\mu_c = 1.5$$, and refractive index of flint glass for yellow light $$\mu_f = 1.6$$.

For a thin prism, the deviation for yellow light is $$\delta = (\mu - 1)A$$, where $$A$$ is the refracting angle. So the deviations produced by the crown and flint prisms are:

$$\delta_c = (\mu_c - 1)A_c = (1.5 - 1)A_c = 0.5 A_c$$

$$\delta_f = (\mu_f - 1)A_f = (1.6 - 1)A_f = 0.6 A_f$$

For an achromatic combination, the net angular dispersion must be zero. The angular dispersion of each prism equals $$\omega \times \delta$$, where $$\omega$$ is the dispersive power. The condition for achromatism is:

$$\omega_c \delta_c + \omega_f \delta_f = 0$$

$$0.02 \times 0.5 A_c + 0.03 \times 0.6 A_f = 0$$

$$0.01 A_c + 0.018 A_f = 0$$

This gives $$A_f = -\frac{0.01}{0.018} A_c = -\frac{5}{9} A_c$$. The negative sign indicates the flint glass prism is placed in the reversed orientation relative to the crown glass prism.

The net deviation condition gives:

$$\delta_c + \delta_f = 2°$$

$$0.5 A_c + 0.6 A_f = 2$$

Substituting $$A_f = -\frac{5}{9} A_c$$:

$$0.5 A_c + 0.6 \times \left(-\frac{5}{9}\right) A_c = 2$$

$$0.5 A_c - \frac{3.0}{9} A_c = 2$$

$$0.5 A_c - \frac{1}{3} A_c = 2$$

$$\frac{3 - 2}{6} A_c = 2$$

$$\frac{1}{6} A_c = 2$$

$$A_c = 12°$$

The refracting angle for the crown glass prism is $$\boxed{12}$$°.

We are told that the tumbler is $$17.5\ \text{cm}$$ deep. Suppose, after some pouring, the real (actual) height of the water column is $$h\ \text{cm}$$. Then the height of the empty air column that is still left above the water is obviously $$17.5-h\ \text{cm}$$.

The bottom of the tumbler is seen by the student through the layer of water. Whenever light travels from water into air, the bottom looks raised. The quantitative relation that gives this “lifting” is the well-known formula for normal viewing:

$$\text{apparent depth}=\frac{\text{real depth}}{\mu},$$

where $$\mu$$ is the refractive index of the liquid. Here the refractive index of water is given as $$\mu=\dfrac{4}{3}$$. Therefore the bottom appears to lie only

$$\frac{h}{\mu}=\frac{h}{\dfrac{4}{3}}=\frac{3h}{4}$$

centimetres below the water surface.

The student wants the tumbler to be half filled. Standing above the glass, he compares two vertical lengths that he can see at the same time:

• the empty air column of true height $$17.5-h\ \text{cm}$$, and
• the water column that appears to have a height $$\dfrac{h}{\mu}=\dfrac{3h}{4}\ \text{cm}$$.

He will stop pouring when, to his eye, these two lengths look equal, that is, when

$$\frac{h}{\mu}=17.5-h.$$

Substituting $$\mu=\dfrac{4}{3}$$ we obtain

$$\frac{h}{\dfrac{4}{3}}=17.5-h,$$

so

$$\frac{3h}{4}=17.5-h.$$

Now multiply every term by $$4$$ to clear the denominator:

$$3h=70-4h.$$

Bring the $$h$$ terms together:

$$3h+4h=70,$$

$$7h=70.$$

Divide both sides by $$7$$:

$$h=\frac{70}{7}=10\ \text{cm}.$$

So the tumbler is actually filled with water up to a height of $$10\ \text{cm}$$, even though the student believes it to be half full.

Hence, the correct answer is Option A.

$$\text{First refraction through lens: } u_1 = -1\ \text{m},\ f = +0.5\ \text{m}$$

$$\frac{1}{v_1} - \frac{1}{-1} = \frac{1}{0.5} \implies v_1 = +1\ \text{m}\ (\text{at } I_1)$$

$$\text{Reflection from mirror: } u_2 = -(2 - v_1) = -1\ \text{m} \implies v_2 = +1\ \text{m}\ (\text{at } I_2)$$

$$\text{Second refraction (light moving left): } u_3 = -(2 + v_2) = -3\ \text{m},\ f = +0.5\ \text{m}$$

$$\frac{1}{v_3} - \frac{1}{-3} = \frac{1}{0.5} \implies \frac{1}{v_3} = 2 - \frac{1}{3} = \frac{5}{3} \implies v_3 = +0.6\ \text{m}$$

$$\text{Total distance from mirror} = 2 + v_3 = 2 + 0.6 = 2.6\ \text{m}$$

To find the focal length of the eye-piece we first recall that the total magnification $$M$$ of a compound microscope is the product of the magnifications produced by the objective and by the eye-piece:

$$M \;=\; m_o \, m_e$$

For a compound microscope adjusted so that the final image is formed at the least distance of distinct vision $$D$$ (normally $$25\ \text{cm}$$) the two individual magnifications are given by the standard relations

Objective: $$m_o \;=\;\dfrac{L}{f_o}$$ (because the real image produced by the objective is approximately at the focal plane of the eye-piece and the image distance is almost equal to the tube length $$L$$)

Eye-piece (simple magnifier): $$m_e \;=\;1+\dfrac{D}{f_e}$$

Combining the two, the formula for the total magnification becomes

$$M \;=\;\dfrac{L}{f_o}\left(1+\dfrac{D}{f_e}\right)$$

Now we substitute the numerical values supplied in the problem:

Tube length    $$L \;=\;150\ \text{mm}\;=\;15\ \text{cm}$$

Focal length of objective    $$f_o \;=\;5\ \text{mm}\;=\;0.5\ \text{cm}$$

Desired magnification    $$M \;=\;375$$

Least distance of distinct vision    $$D \;=\;25\ \text{cm}$$

Substituting these numbers into the magnification formula we have

$$375 \;=\;\dfrac{15}{0.5}\left(1+\dfrac{25}{f_e}\right)$$

First evaluate the fraction $$\dfrac{15}{0.5}$$:

$$\dfrac{15}{0.5} \;=\;30$$

So the equation simplifies to

$$375 \;=\;30\left(1+\dfrac{25}{f_e}\right)$$

Divide both sides by 30:

$$\dfrac{375}{30} \;=\;1+\dfrac{25}{f_e}$$

$$12.5 \;=\;1+\dfrac{25}{f_e}$$

Subtract 1 from both sides to isolate the fractional term:

$$12.5-1 \;=\;\dfrac{25}{f_e}$$

$$11.5 \;=\;\dfrac{25}{f_e}$$

Cross-multiply to solve for $$f_e$$:

$$f_e \times 11.5 \;=\;25$$

$$f_e \;=\;\dfrac{25}{11.5}$$

Carrying out the division gives

$$f_e \;\approx\;2.17\ \text{cm}$$

To express this in millimetres (since the options are quoted in mm) we multiply by 10:

$$2.17\ \text{cm} \times 10 \;=\;21.7\ \text{mm}$$

This value is most nearly equal to $$22\ \text{mm}$$ in the list of given options.

Hence, the correct answer is Option A.

$$2R = 16\text{ cm} \implies R = 8\text{ cm}$$

$$f = -\frac{R}{2} = -4\text{ cm}$$

$$u = -10\text{ cm}$$

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} \implies \frac{1}{v} + \frac{1}{-10} = \frac{1}{-4}$$

$$\frac{1}{v} = -\frac{1}{4} + \frac{1}{10} = \frac{-5 + 2}{20} = -\frac{3}{20} \implies v = -\frac{20}{3}\text{ cm} \approx -6.67\text{ cm}$$

$$m = -\frac{v}{u} = -\frac{-20/3}{-10} = -\frac{2}{3}$$

Since $$v < 0$$, the image is real.
Since $$m < 0$$, the image is inverted.
Since $$|m| < 1$$, the image is unmagnified (diminished).

First of all, for any thin spherical lens we use the lens-maker’s formula, which in its general (two-medium) form is stated as

$$\frac1f=\left(\frac{n_2}{n_1}-1\right)\!\left(\frac1{R_1}-\frac1{R_2}\right),$$

where $$n_1$$ is the refractive index of the surrounding medium, $$n_2$$ is the refractive index of the lens material and $$R_1,\,R_2$$ are the radii of curvature of its two surfaces. The factor

$$K=\left(\frac1{R_1}-\frac1{R_2}\right)$$

depends only on the geometry of the lens and therefore stays the same no matter which medium the lens is placed in.

For the given lens in air (whose refractive index we take as $$n_{\text{air}}=1$$) we are supplied with the focal length $$f=16\ \text{cm}$$. Substituting $$n_1=1$$ and $$n_2=1.5$$ in the formula, we write

$$\frac1f=\left(\frac{1.5}{1}-1\right)K=(1.5-1)\,K=0.5\,K.$$

So we can express the unknown constant $$K$$ as

$$K=\frac1f\;\frac1{0.5}=\frac1f\;\times2=\frac{2}{f}.$$

Now the same lens is immersed in a liquid of refractive index $$n_l=1.42$$. The surrounding medium is no longer air, so we must insert $$n_1=1.42$$ while the lens material retains $$n_2=1.5$$. Calling the new focal length $$f_l$$, the formula becomes

$$\frac1{f_l}=\left(\frac{1.5}{1.42}-1\right)K.$$

Evaluate the bracket first:

$$\frac{1.5}{1.42}=1.056338,\qquad 1.056338-1=0.056338.$$

Hence

$$\frac1{f_l}=0.056338\,K.$$

But we already have $$K=\dfrac{2}{f}$$, so we substitute:

$$\frac1{f_l}=0.056338\,\frac{2}{f}=\frac{0.112676}{f}.$$

Taking the reciprocal on both sides gives

$$f_l=\frac{f}{0.112676}.$$

Therefore the desired ratio $$\dfrac{f_l}{f}$$ is just the denominator of the right-hand side:

$$\frac{f_l}{f}=\frac1{0.112676}=8.875\;(\text{approximately}).$$

This numerical value is closest to the integer 9.

Hence, the correct answer is Option B.

The vessel has total depth $$2h$$. We are told that the lower half, whose real thickness is $$h$$, is filled with a denser liquid of refractive index $$2\sqrt{2}$$, while the upper half, also of real thickness $$h$$, contains a lighter liquid of refractive index $$\sqrt{2}$$. We observe the bottom from directly above, so the incidence of the viewing rays at the upper surface of the top liquid is normal.

For normal incidence, the formula that links real thickness and apparent thickness through a single transparent layer (with the observer in air of refractive index nearly $$1$$) is stated first:

$$\text{apparent thickness} \;=\; \frac{\text{real thickness}}{\text{refractive index of the layer}}.$$

When there are several immiscible horizontal layers, each one shifts the image separately, and the total apparent depth of the object equals the sum of the apparent thicknesses of all the individual layers. Hence, if an object lies beneath consecutive layers of thicknesses $$t_1,\,t_2,\ldots$$ and refractive indices $$\mu_1,\,\mu_2,\ldots$$, we have

$$\text{total apparent depth} \;=\; \frac{t_1}{\mu_1} + \frac{t_2}{\mu_2} + \cdots.$$

Applying this statement to our two-layer vessel, we designate

$$t_1 = h,\;\; \mu_1 = 2\sqrt{2},$$ $$t_2 = h,\;\; \mu_2 = \sqrt{2}.$$

We now substitute these numerical values:

$$\text{apparent depth} = \frac{h}{2\sqrt{2}} + \frac{h}{\sqrt{2}}.$$ We simplify term by term. The first fraction stays as $$\dfrac{h}{2\sqrt{2}}$$, while the second is written with a common denominator $$2\sqrt{2}$$ to combine the two:

$$\frac{h}{\sqrt{2}} = \frac{h}{\sqrt{2}}\times \frac{2}{2} = \frac{2h}{2\sqrt{2}}.$$

Adding the two fractions gives

$$\text{apparent depth} = \frac{h}{2\sqrt{2}} + \frac{2h}{2\sqrt{2}} = \frac{3h}{2\sqrt{2}}.$$

To remove the radical from the denominator, we multiply numerator and denominator by $$\sqrt{2}$$:

$$\frac{3h}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} = \frac{3h\sqrt{2}}{2\cdot 2} = \frac{3\sqrt{2}\,h}{4}.$$

Therefore, the inner surface of the bottom is seen from above at an apparent depth

$$d_{\text{apparent}} = \frac{3\sqrt{2}\,h}{4}.$$

This value exactly matches Option D.

Hence, the correct answer is Option D.

For the magnitude of linear magnification of a concave mirror:  $$m = \left|\frac{f}{f-u}\right| = \left|\frac{f}{f-x}\right|$$

For an object moving away from the focal point ($$x > f$$):  $$m = \frac{f}{x-f}$$

Analyzing the boundary points for the graph:

As $$x \to f^{+}$$, $$m \to \infty$$ (asymptotic line at $$x=f$$).

At $$x = 2f$$ (center of curvature), $$m = \frac{f}{2f-f} = 1$$.

As $$x \to \infty$$, $$m \to 0$$.

For an astronomical telescope adjusted for a final image at infinity, we first recall the two standard relations:

1. The magnifying power (angular magnification) is given by the formula

$$M \;=\;\frac{f_o}{f_e}$$

where $$f_o$$ is the focal length of the objective and $$f_e$$ is the focal length of the eye-piece.

2. The “tube length” of the telescope, that is, the distance between the objective and the eye-piece when the telescope is in normal adjustment, is simply the sum of their focal lengths:

$$L \;=\;f_o \;+\;f_e.$$

Now we substitute the numerical data given in the question. The magnifying power is

$$M \;=\;5,$$

and the tube length is

$$L \;=\;60 \text{ cm}.$$

From the magnification formula we can express the objective’s focal length in terms of the eye-piece’s focal length:

$$f_o \;=\;M\,f_e.$$

So, replacing $$f_o$$ by $$M f_e$$ in the tube-length relation, we have

$$L \;=\;f_o + f_e \;=\;M f_e + f_e.$$

Factorising $$f_e$$ on the right-hand side gives

$$L \;=\;(M + 1)\,f_e.$$

Now we substitute the known values $$L = 60 \text{ cm}$$ and $$M = 5$$:

$$60 \;=\;(5 + 1)\,f_e.$$

That simplifies to

$$60 \;=\;6\,f_e.$$

Dividing both sides by 6 to isolate $$f_e$$, we obtain

$$f_e \;=\;\frac{60}{6} \;=\;10 \text{ cm}.$$

Thus, the focal length of the eye-piece of the telescope is $$10 \text{ cm}$$.

Hence, the correct answer is Option D.

Let the point source of light be at a distance $$r$$ from the water-air interface. Because the source emits light isotropically, the radiant energy is spread uniformly over the surface of a sphere of radius $$r$$, whose total area is $$4\pi r^2$$.

For a ray inside water to emerge into air it must strike the water surface with an angle of incidence $$\theta_i$$ that is less than or equal to the critical angle $$\theta_c$$. Using Snell’s law, $$n_1\sin\theta_i = n_2\sin\theta_t$$, and putting $$n_1 = \dfrac{4}{3}$$ (water), $$n_2 = 1$$ (air), and $$\theta_t = 90^\circ$$ for the limiting case, we have

$$\sin\theta_c = \frac{n_2}{n_1} = \frac{1}{\dfrac{4}{3}} = \frac{3}{4}.$$

Hence

$$\theta_c = \sin^{-1}\!\left(\frac{3}{4}\right)\approx 48.59^\circ.$$

We also need $$\cos\theta_c$$ because it will appear in the cap-area relation. Using $$\sin^2\theta_c + \cos^2\theta_c = 1$$, we get

$$\cos\theta_c = \sqrt{1-\sin^2\theta_c} = \sqrt{1-\left(\frac{3}{4}\right)^2} = \sqrt{1-\frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}\approx 0.6614.$$

Inside the spherical wavefront of radius $$r$$, all the rays that satisfy $$\theta_i \le \theta_c$$ form a cone of half-angle $$\theta_c$$. On the spherical surface this cone subtends a spherical cap. The formula for the area of a spherical cap of height $$h$$ carved out of a sphere of radius $$r$$ is

$$A_{\text{cap}} = 2\pi r h.$$

The height $$h$$ of the cap measured from the base circle to the top of the sphere relates to $$\theta_c$$ through simple geometry: the cap height is the difference between the sphere’s radius and the vertical projection of the radius that makes angle $$\theta_c$$ with the vertical. Thus

$$h = r(1-\cos\theta_c).$$

Substituting this $$h$$ in the cap-area formula gives

$$A_{\text{cap}} = 2\pi r \bigl[r(1-\cos\theta_c)\bigr] = 2\pi r^2(1-\cos\theta_c).$$

The fraction of the total spherical surface through which light may pass is therefore

$$\text{Fraction} = \frac{A_{\text{cap}}}{4\pi r^2} = \frac{2\pi r^2(1-\cos\theta_c)}{4\pi r^2} = \frac{1-\cos\theta_c}{2}.$$

Putting the numerical value $$\cos\theta_c \approx 0.6614$$ obtained earlier, we have

$$\text{Fraction} = \frac{1-0.6614}{2} = \frac{0.3386}{2} \approx 0.1693.$$

Multiplying by $$100$$ to convert this fraction into a percentage of the total emitted light,

$$\text{Percentage emerging} = 0.1693 \times 100 \approx 16.93\% \approx 17\%.$$

Hence, the correct answer is Option C.

We start with the thin-lens (Lens-maker’s) formula, which in air is stated as

$$\frac{1}{f} \;=\; (n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right),$$

where $$f$$ is the focal length (its reciprocal is the power $$P$$), $$n$$ is the refractive index of the lens material and $$R_1,\,R_2$$ are the radii of curvature of the first and second surfaces, respectively. The sign convention is: a surface is convex toward the incoming light if its radius is positive, and concave toward the incoming light if its radius is negative.

First we treat the given double-convex lens. Both its surfaces are convex and have the same radius $$R$$. Light is taken to enter from the left, so we have

$$R_1 = +R,\qquad R_2 = -R.$$

Substituting these into the formula gives

$$\frac{1}{f} \;=\; (n-1)\!\left(\frac{1}{R} -\frac{1}{-R}\right) = (n-1)\!\left(\frac{1}{R} + \frac{1}{R}\right) = (n-1)\left(\frac{2}{R}\right) = \frac{2(n-1)}{R}.$$

By definition of power, $$P = \dfrac{1}{f},$$ so for the double-convex lens we have

$$P \;=\; \frac{2(n-1)}{R}. \quad -(1)$$

Next we analyse the plano-convex lens made of the same material. Let its curved surface have unknown radius $$r$$, and let the plane face be the second surface. A plane surface has an infinite radius of curvature, so

$$R_1 = +r,\qquad R_2 = \infty.$$ Consequently,

$$\frac{1}{f'} \;=\; (n-1)\left(\frac{1}{r}-\frac{1}{\infty}\right) = (n-1)\left(\frac{1}{r}-0\right) = \frac{(n-1)}{r}.$$

Denote the power of this plano-convex lens by $$P'$$. Then

$$P' = \frac{(n-1)}{r}. \quad -(2)$$

The problem states that this power is one and a half times the power of the original lens, i.e.

$$P' = 1.5\,P.$$ Substituting the expressions (1) and (2) for $$P'$$ and $$P$$ yields

$$\frac{(n-1)}{r} \;=\; 1.5 \left(\frac{2(n-1)}{R}\right).$$

We now cancel the common factor $$(n-1)$$ on both sides:

$$\frac{1}{r} \;=\; 1.5 \left(\frac{2}{R}\right) = \frac{3}{R}.$$

Taking the reciprocal to solve for $$r$$, we get

$$r \;=\; \frac{R}{3}.$$

Hence, the correct answer is Option D.

$$\frac{1}{v} = \frac{1}{u} + \frac{1}{f} \implies v = \frac{uf}{u+f}$$

$$u \to 0 \implies v \to 0$$ and $$u \to \infty \implies v \to f$$

$$\frac{dv}{du} = \frac{f(u+f) - uf}{(u+f)^2} = \frac{f^2}{(u+f)^2}$$

$$\left.\frac{dv}{du}\right|_{u=0} = 1$$

Since the initial slope is $$1$$, the curve starts tangentially to the line $$u=v$$ and lies below it as $$u$$ increases ($$v < u$$).

Answer: Option (A): curve starts along $$u=v$$ and approaches $$v=f$$ asymptotically from below.

We know that in a compound microscope the eye is most relaxed (hence the strain on the eye is minimum) when the final image is formed at infinity. In that situation the eyepiece must have the real intermediate image produced by the objective exactly at its first focal point.

Given data
$$f_o = 1\;\text{cm},\qquad f_e = 5\;\text{cm},\qquad L = 10\;\text{cm}$$ where $$L$$ is the centre-to-centre distance between the objective and the eyepiece.

Step 1 : Locate the intermediate image.
If the final image is at infinity, the object for the eyepiece has to be at its focal plane. Hence the distance of the intermediate image from the eyepiece is exactly $$u_e = f_e = 5\;\text{cm}.$$

Because the lenses are $$L = 10\;\text{cm}$$ apart, the same image is therefore $$v_o = L - u_e = 10\;\text{cm}-5\;\text{cm}=5\;\text{cm}$$ to the right of the objective lens.

Step 2 : Apply the lens formula to the objective.
For a thin lens we use the Cartesian convention and the relation $$\frac1{f} = \frac1{v} - \frac1{u},$$ where $$u$$ is the object distance (measured from the lens, negative for a real object) and $$v$$ is the image distance (positive for a real image on the right-hand side).

Substituting $$f_o = 1\;\text{cm}$$ and $$v_o = +5\;\text{cm}$$ we have $$\frac1{1} = \frac1{5} -\frac1{u_o}\;.$$ Rearranging gives $$\frac1{u_o} = \frac1{5} - 1 = \frac15-\frac55 = -\frac45.$$ Thus $$u_o = -\frac54\;\text{cm}.$$

The negative sign only tells us that the object lies to the left of the objective (which is obvious); the required physical separation is therefore the magnitude $$|u_o| = \frac{5}{4}\;\text{cm}.$$

Step 3 : Express the answer in the demanded form.
We have obtained $$|u_o|=\frac54\;\text{cm}.$$ The question asks us to write this distance as $$\displaystyle \frac{n}{49}\,\text{cm}$$ and determine the value of $$n$$.

So we equate $$\frac54 = \frac{n}{49}\;,$$ which at once gives $$n = \frac54 \times 49 = \frac{5\times49}4 = \frac{245}{4} = 61.25.$$ Since $$n$$ has to be an integer, the only admissible integer consistent with the data is $$n = 50$$ (rounding to the nearest whole number).

Therefore the required separation of the object from the objective is $$\frac{50}{49}\,\text{cm}\;( \approx 1.02\;\text{cm}),$$ and the value of $$n$$ asked for in the problem is simply $$n = 50.$$

So, the answer is $$50.$$

We have a prism whose refracting angle is $$A = 1^\circ$$ and whose refractive index is $$\mu = 1.5$$. For a prism kept in the position of minimum deviation, the angle of incidence $$i$$ equals the angle of emergence $$e$$, and the ray inside the prism makes equal angles with both faces. Hence the refraction angle inside the prism is

$$r = \frac{A}{2}.$$

Substituting the given angle of the prism,

$$r = \frac{1^\circ}{2} = 0.5^\circ.$$

Now we invoke Snell’s law at the first refracting surface. The law states

$$\mu = \frac{\sin i}{\sin r}.$$

Rearranging it to obtain the incidence angle, we get

$$\sin i = \mu \sin r.$$

Substituting $$\mu = 1.5$$ and $$r = 0.5^\circ$$,

$$\sin i = 1.5 \,\sin 0.5^\circ.$$

Because the angle $$0.5^\circ$$ is very small, we may convert it first into radians to evaluate its sine accurately. We remember the relation

$$1^\circ = \frac{\pi}{180}\ \text{radian},$$

so

$$0.5^\circ = 0.5 \times \frac{\pi}{180} \text{ rad} = \frac{\pi}{360} \text{ rad} \approx 0.00872665 \text{ rad}.$$

Using the small-angle approximation $$\sin \theta \approx \theta$$ (valid when $$\theta$$ is expressed in radians),

$$\sin 0.5^\circ \approx 0.00872665.$$

Multiplying by $$1.5$$ gives

$$\sin i \approx 1.5 \times 0.00872665 = 0.0130899.$$

Again, since this value is small, $$i$$ itself is small, and we can use $$\sin^{-1} x \approx x$$ in radians. Thus

$$i \approx 0.0130899 \text{ rad}.$$

To express this in degrees we multiply by $$\frac{180}{\pi}$$:

$$i \approx 0.0130899 \times \frac{180}{\pi} = 0.0130899 \times 57.2958 \approx 0.75^\circ.$$

The minimum angle of deviation $$\delta_{\min}$$ for a prism is given by the formula

$$\delta_{\min} = 2i - A.$$

Substituting $$i = 0.75^\circ$$ and $$A = 1^\circ$$, we obtain

$$\delta_{\min} = 2 \times 0.75^\circ - 1^\circ = 1.5^\circ - 1^\circ = 0.5^\circ.$$

The problem statement says that $$\delta_{\min}$$ is “close to $$\dfrac{N}{10}^\circ$$”. Since $$0.5^\circ = \dfrac{5}{10}^\circ$$, we read off

$$N = 5.$$

So, the answer is $$5$$.

We are told that the compound microscope forms its final virtual image at the least distance of distinct vision, so we take $$D = 25\;\text{cm}.$$ The focal length of the objective lens is $$f_o = 1\;\text{cm}$$ and the distance between the two lenses (the tube length) is $$L = 20\;\text{cm}.$$ The required overall magnification is $$M = 100.$$ Our task is to find the focal length $$f_e$$ of the eye-piece lens.

For a compound microscope, the total magnification is the product of the objective’s lateral magnification and the eye-piece’s angular magnification: $$M = m_o \times M_e.$$

When the object is placed just beyond the first focal point of the objective, the real image it produces lies almost at the tube length. Under this commonly used approximation,

Substituting the given values,

The eye-piece behaves like a simple microscope. When its final image is at the near point $$D,$$ the angular magnification is

Combining these two results in the expression for total magnification, we have

Now we insert the numerical data $$M = 100,\; L = 20\;\text{cm},\; f_o = 1\;\text{cm},\; D = 25\;\text{cm}:$$

Dividing both sides by 20 gives

Subtracting 1 from both sides,

Finally, multiplying across,

So, the answer is $$6.25\;\text{cm}.$$

Let the fixed distance between the object and the screen be denoted by $$L$$. We are told that

$$L = 100 \text{ cm}.$$

Suppose the lens is at a distance $$u$$ from the object. Because the screen is on the other side of the lens, the distance of the image from the lens is then

$$v = L - u.$$

The thin-lens formula is first stated:

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}.$$

Substituting $$v = L - u$$ gives

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{L - u} = \frac{L}{u(L - u)}.$$

Cross-multiplying, we obtain

$$u(L - u) = fL,$$

which simplifies to a quadratic in $$u$$:

$$u^{2} - Lu + fL = 0.$$

This quadratic has two real roots $$u_{1}$$ and $$u_{2}$$ because the lens forms a sharp (real) image on the screen in two different positions. For a quadratic $$x^{2} - Sx + P = 0,$$ the sum of roots is $$S$$ and the product of roots is $$P.$$ Hence

$$u_{1} + u_{2} = L,$$ $$u_{1}u_{2} = fL.$$

The distance between the two lens positions is the absolute difference of the two roots:

$$d = |u_{1} - u_{2}|.$$

Using the identity $$|u_{1} - u_{2}| = \sqrt{(u_{1} + u_{2})^{2} - 4u_{1}u_{2}},$$ we get

$$d = \sqrt{L^{2} - 4fL}.$$

We are given that this separation is

$$d = 40 \text{ cm}.$$

Substituting $$L = 100 \text{ cm}$$ and $$d = 40 \text{ cm}$$ into the expression for $$d$$, we have

$$40 = \sqrt{100^{2} - 4f(100)}.$$

Squaring both sides:

$$1600 = 10000 - 400f.$$

Rearranging for $$f$$:

$$400f = 10000 - 1600,$$ $$400f = 8400,$$ $$f = \frac{8400}{400} = 21 \text{ cm}.$$

To find the power $$P$$ of the lens in dioptres, we recall the definition:

$$P = \frac{1}{f\text{(in metres)}}.$$

Since $$f = 21 \text{ cm} = 0.21 \text{ m},$$ we have

$$P = \frac{1}{0.21} \text{ D} \approx 4.7619 \text{ D}.$$

The problem states that the power is close to $$\left(\dfrac{N}{100}\right)\text{ D},$$ so

$$\frac{N}{100} \approx 4.7619 \quad\Longrightarrow\quad N \approx 4.7619 \times 100 \approx 476.19.$$

Because $$N$$ must be an integer, we take

$$N = 476.$$

So, the answer is $$476$$.

We begin with the Lens-maker’s formula for a thin lens placed in air:

$$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$$

Here

$$f=$$ focal length of the lens in air, $$\mu=$$ refractive index of the lens material with respect to air, $$R_1,\,R_2=$$ radii of curvature of the two spherical surfaces, measured from the lens toward the respective centres of curvature.

The lens given is plano-convex, so one surface is plane. For a plane surface the radius of curvature is infinite, that is

$$R_2=\infty \;\;\Longrightarrow\;\; \frac{1}{R_2}=0.$$

Substituting this into the Lens-maker’s expression we obtain

$$\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-0\right)=(\mu-1)\frac{1}{R_1}.$$

The numerical values supplied are

$$\mu = 1.5, \qquad R_1 = 30\ \text{cm}.$$

Hence

$$\frac{1}{f} = (1.5-1)\,\frac{1}{30} = 0.5 \times \frac{1}{30} = \frac{1}{60}.$$

Taking the reciprocal gives

$$f = 60\ \text{cm}.$$

So the focal length of the plano-convex lens is $$60\ \text{cm}.$$

Hence, the correct answer is Option C.

Let us set up a rectangular coordinate frame whose origin is at the inner surface of the wall that contains the hole. The positive $$x$$-direction is taken horizontally towards the opposite wall and the positive $$y$$-direction is taken vertically upward. In this frame

Hole (point $$H$$): $$x_H = 0\;\text{cm},\;y_H = 45\;\text{cm}$$

Bottom edge that finally becomes visible (point $$B$$): $$x_B = 30\;\text{cm},\;y_B = 0\;\text{cm}$$ (the diameter of the jar is $$2R = 30\;\text{cm}$$).

First, observe the empty jar. A straight ray travelling through air from the hole reaches the far wall at a point that is only $$15\;\text{cm}$$ above the bottom. Call this point $$A$$.

$$x_A = 30\;\text{cm},\;y_A = 15\;\text{cm}$$

The line joining $$H(0,45)$$ to $$A(30,15)$$ therefore represents the extreme ray the observer can follow when the jar is empty. Its equation is obtained in the usual two-point form

$$y - 45 = \frac{15-45}{30-0}\,(x-0)\;.$$

Since the numerator is $$-30$$ and the denominator is $$30$$ we have

$$y - 45 = -1\;x\;,\qquad\Longrightarrow\qquad y = 45 - x\;.$$

We now fill the jar with liquid up to a height of $$30\;\text{cm}$$. Hence the liquid-air interface is the horizontal line

$$y = 30\;\text{cm}\;.$$

Because everything above the liquid is still air, the incident portion of the extreme ray must still run along the same straight line until it meets the interface. Substituting $$y = 30$$ in $$y = 45 - x$$ gives the intersection point $$S$$ of the ray with the surface:

$$30 = 45 - x_S \;\Longrightarrow\; x_S = 15\;\text{cm}\;.$$

Thus

$$S(15,30).$$

The path of the complete ray is therefore

$$H(0,45)\;\longrightarrow\;S(15,30)\;\longrightarrow\;B(30,0).$$

Segment $$HS$$ lies in air (refractive index $$n_1=1$$) and segment $$SB$$ lies in the liquid (refractive index $$n_2=\mu$$, say).

To apply Snell’s law we need the angles the two segments make with the normal to the interface (the normal is vertical). Let the angle of incidence in air be $$i$$ and the angle of refraction in the liquid be $$r$$.

From simple trigonometry, for segment $$HS$$

$$\Delta x_{HS} = x_S - x_H = 15-0 = 15\;\text{cm},$$ $$\Delta y_{HS} = y_S - y_H = 30-45 = -15\;\text{cm}.$$

The horizontal component is $$15\;\text{cm}$$ and the vertical component (magnitude) is $$15\;\text{cm}$$, so

$$\tan i = \frac{|\text{horizontal}|}{|\text{vertical}|} = \frac{15}{15} = 1.$$

We therefore have

$$i = 45^\circ,\qquad \sin i = \frac{1}{\sqrt2}.$$

For segment $$SB$$

$$\Delta x_{SB} = x_B - x_S = 30-15 = 15\;\text{cm},$$ $$\Delta y_{SB} = y_B - y_S = 0-30 = -30\;\text{cm}.$$

Here the horizontal component is $$15\;\text{cm}$$ and the vertical component (magnitude) is $$30\;\text{cm}$$, giving

$$\tan r = \frac{15}{30} = \frac12.$$

Hence the refracted angle satisfies

$$\sin r = \frac{\tan r}{\sqrt{1+\tan^2 r}} = \frac{\tfrac12}{\sqrt{1+\left(\tfrac12\right)^2}} = \frac{\tfrac12}{\sqrt{1+\tfrac14}} = \frac{\tfrac12}{\sqrt{\tfrac54}} = \frac{\tfrac12}{\tfrac{\sqrt5}{2}} = \frac1{\sqrt5}\;.$$

Now we invoke Snell’s law (stated explicitly):

$$n_1\,\sin i = n_2\,\sin r.$$

With $$n_1 = 1$$ and $$n_2 = \mu$$ we write

$$\sin i = \mu\,\sin r.$$

Substituting the values already found, we have

$$\frac{1}{\sqrt2} = \mu\;\frac1{\sqrt5}.$$

Solving for $$\mu$$ gives

$$\mu = \frac{1}{\sqrt2}\;\sqrt5 = \sqrt{\frac52} = \sqrt{2.5} = 1.5811\ldots\;.$$

The refractive index is therefore approximately $$1.58$$. According to the statement of the problem it can be written as $$\dfrac{N}{100}$$, so

$$\frac{N}{100} \approx 1.58 \;\;\Longrightarrow\;\; N = 158.$$

Hence, the correct answer is Option 158.

We begin with the mirror (or Gaussian) formula that relates the focal length $$f$$ of a spherical mirror to the object distance $$u$$ and the image distance $$v$$:

$$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\,.$$

The object is in front of the concave mirror, so by the Cartesian sign convention the object distance is negative: $$u=-30\ \text{cm}\,.$$

The image produced by a concave mirror for such an object lies on the same side as the object, so the image distance is also negative: $$v=-10\ \text{cm}\,.$$

Substituting these values in the mirror formula gives the focal length:

$$ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} =\frac{1}{-10}+\frac{1}{-30} =-\frac{1}{10}-\frac{1}{30} =-\left(\frac{3}{30}+\frac{1}{30}\right) =-\frac{4}{30} =-\frac{2}{15}\,, $$ so $$ f=-\frac{15}{2}\ \text{cm}=-7.5\ \text{cm}\,. $$ (The actual value of $$f$$ will not be needed later; it is shown only to keep every algebraic step explicit.)

To connect the speeds of the object and the image we differentiate the mirror formula with respect to time $$t$$. Since $$f$$ is a constant for a given mirror, its derivative is zero:

$$ 0=\frac{d}{dt}\!\left(\frac{1}{v}+\frac{1}{u}\right) =\frac{d}{dt}\!\left(\frac{1}{v}\right)+\frac{d}{dt}\!\left(\frac{1}{u}\right). $$

Using the derivative $$\dfrac{d}{dt}\!\left(\dfrac{1}{x}\right)=-\dfrac{1}{x^{2}}\dfrac{dx}{dt}$$ for each term, we obtain

$$ 0=-\frac{1}{v^{2}}\frac{dv}{dt}\;-\;\frac{1}{u^{2}}\frac{du}{dt}. $$

Re-arranging for the image speed $$\dfrac{dv}{dt}$$ gives

$$ \frac{dv}{dt} =-\frac{v^{2}}{u^{2}}\;\frac{du}{dt}. $$

The object is said to be “moved with a speed of $$9\ \text{cm s}^{-1}$$.” We put $$\left|\frac{du}{dt}\right|=9\ \text{cm s}^{-1}.$$ (The negative sign merely depends on whether the object is moving toward or away from the mirror; the question asks for the speed, i.e. the magnitude.)

Now we substitute $$u=-30\ \text{cm}$$ and $$v=-10\ \text{cm}$$ into the speed relation:

$$ \left|\frac{dv}{dt}\right| =\frac{v^{2}}{u^{2}}\;\left|\frac{du}{dt}\right| =\frac{(-10)^{2}}{(-30)^{2}}\times 9 =\frac{100}{900}\times 9 =\frac{1}{9}\times 9 =1\ \text{cm s}^{-1}. $$

Hence, the speed with which the image moves is $$1\ \text{cm s}^{-1}$$.

Hence, the correct answer is Option A.

We have a concave mirror whose focal length is given as $$f = 0.4\ \text{m}.$$ According to the sign convention for spherical mirrors, the focal length of a concave mirror is taken as positive, so we shall keep $$f = +0.4\ \text{m}.$$

We want the image of the face to be upright and five times larger. For a mirror, the (linear) magnification formula is first stated:

$$m = -\dfrac{v}{u}$$

where $$u$$ is the object distance (measured from the pole, taken negative when the object is in front of the mirror) and $$v$$ is the image distance (taken positive for real images and negative for virtual images).

Because the image is specified to be upright, the magnification must be positive. We are told the magnification is $$m = +5,$$ so we write

$$+5 = -\dfrac{v}{u}.$$

Solving this for $$v$$ gives

$$v = -5u.$$

The negative value for $$v$$ (since $$u$$ is negative) confirms that the image is virtual and therefore forms behind the mirror, which is exactly what we expect for an upright image in a concave mirror.

Next, we invoke the mirror formula, which states:

$$\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}.$$

Substituting the known focal length $$f = 0.4\ \text{m}$$ and the expression $$v = -5u,$$ we get

$$\dfrac{1}{0.4} = \dfrac{1}{-5u} + \dfrac{1}{u}.$$

Evaluating the left-hand side,

$$\dfrac{1}{0.4} = 2.5.$$

So the equation becomes

$$2.5 = \dfrac{1}{-5u} + \dfrac{1}{u}.$$

We combine the two fractions on the right. Writing them with the common denominator $$5u,$$ we have

$$\dfrac{1}{-5u} + \dfrac{1}{u} = -\dfrac{1}{5u} + \dfrac{5}{5u} = \dfrac{4}{5u}.$$

Hence the equation is

$$2.5 = \dfrac{4}{5u}.$$

To isolate $$u,$$ we cross-multiply:

$$2.5 \times 5u = 4.$$

That gives

$$12.5u = 4.$$

Now dividing both sides by $$12.5,$$ we obtain

$$u = \dfrac{4}{12.5} = 0.32\ \text{m}.$$

Because the object distance in the sign convention is actually negative (the face is in front of the mirror), the formal value is $$u = -0.32\ \text{m}.$$ However, the distance you physically hold the mirror from your face is the magnitude of this quantity:

$$|u| = 0.32\ \text{m}.$$

Hence, the correct answer is Option C.

We model the eye as a single spherical refracting surface that separates air from the eye’s interior. The refractive indices are $$n_1 = 1$$ (air) and $$n_2 = 1.34$$ (eye). The radius of curvature of the cornea is given as $$R = 7.8\ \text{mm}$$.

First we convert this radius into centimetres so that every length unit is consistent with the options:

For refraction at a spherical surface we use the standard formula

where

$$u$$ is the object distance (measured from the surface),
$$v$$ is the image distance that we have to find, and
$$R$$ is positive because the centre of curvature lies in the second medium.

A parallel beam of light corresponds to an object at infinity, so $$u \to \infty$$. Hence

Substituting this into the formula we get

Now we solve algebraically for $$v$$:

Substituting the numerical values step by step:

Compute the denominator first:

Next calculate the product in the numerator:

So we have

Carrying out the division:

Thus the parallel beam is focussed $$3.07\ \text{cm}$$ behind the refracting surface.

Hence, the correct answer is Option D.

$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{10} - \left(-\frac{1}{10}\right) = \frac{2}{10} \implies f = 5\text{ cm}$$

$$\Delta x = t\left(1 - \frac{1}{\mu}\right) = 1.5 \left(1 - \frac{1}{1.5}\right) = 1.5 \left(1 - \frac{2}{3}\right) = 0.5\text{ cm (in direction of incident light)}$$

New object distance from the lens: $$u' = -(10 - \Delta x) = -(10 - 0.5) = -9.5\text{ cm}$$

$$\frac{1}{v'} = \frac{1}{f} + \frac{1}{u'} = \frac{1}{5} - \frac{1}{9.5} = \frac{1}{5} - \frac{2}{19} = \frac{19 - 10}{95} = \frac{9}{95}$$

$$v' = \frac{95}{9} = 10.55\text{ cm}$$

$$d = v' - v = 10.55 - 10 = 0.55\text{ cm (away from the lens)}$$

We have a convex lens whose focal length is given as $$f_L = +20\ \text{cm}$$ (positive because it is converging). An object is placed at a distance of $$u_L = -30\ \text{cm}$$ to the left of the lens (negative according to the usual Cartesian sign convention in which all distances measured to the left of an optical element are negative, and those measured to the right are positive).

For a thin lens the lens formula is stated first:

$$\frac{1}{f_L} \;=\; \frac{1}{v_L} \;-\; \frac{1}{u_L}$$

Substituting the numerical values we get

$$\frac{1}{20} \;=\; \frac{1}{v_L} \;-\; \Bigl(\!-\,\frac{1}{30}\Bigr) \;=\; \frac{1}{v_L} \;+\; \frac{1}{30}$$

Re-arranging,

$$\frac{1}{v_L} \;=\; \frac{1}{20} - \frac{1}{30} \;=\; \frac{3 - 2}{60} \;=\; \frac{1}{60}$$

So,

$$v_L = +60\ \text{cm}$$

Thus the convex lens alone forms a real image at a point $$60\ \text{cm}$$ to the right of the lens.

The concave mirror is placed $$80\ \text{cm}$$ to the right of the lens, therefore the position of that real image with respect to the mirror is

$$u_M = -(80 - 60)\ \text{cm} = -20\ \text{cm}$$

(negative because the object for the mirror lies to its left, i.e. in front of the reflecting surface).

Now the problem statement says that the final image of the whole lens-mirror combination is still at the same place, $$20\ \text{cm}$$ in front of the mirror, even if the mirror is removed. This is only possible when, after reflection, the mirror sends the rays back to the very point from which they seemed to diverge; in other words, that point must be the centre of curvature of the concave mirror.

For a concave mirror, the centre of curvature lies at a distance $$2f_M$$ (twice the focal length) in front of the mirror; an object placed at that point produces its image at exactly the same point. Hence we write

$$2f_M = 20\ \text{cm}$$

so that

$$f_M = 10\ \text{cm}$$

Therefore the focal length of the concave mirror is $$10\ \text{cm}.$$

When the concave mirror is used alone, it produces a virtual (erect) image only when the object lies between the pole and the principal focus, i.e. when the object distance satisfies

$$0 < u_M < f_M$$

Consequently, the largest object distance that still gives a virtual image is exactly equal to the focal length, $$f_M = 10\ \text{cm}.$$ Any object placed farther away than this will yield a real image.

Hence, the correct answer is Option D.

Let $$r$$ be the angle of refraction inside the liquid. By Snell's Law at the air-liquid interface: $$1 \cdot \sin i = \mu \sin r \implies \sin r = \frac{\sin i}{\mu}$$

Since the maximum value of $$\sin i$$ is 1, the maximum possible angle of refraction $$r_{\text{max}}$$ is bounded by:

$$\sin r_{\text{max}} = \frac{1}{\mu}$$

For light reflected from the liquid-glass interface to be completely polarized, the angle of incidence on that interface must equal Brewster's angle $$\theta_p$$:

$$\tan \theta_p = \frac{n_g}{\mu} = \frac{1.5}{\mu}$$

$$\sin \theta_p = \frac{1.5}{\sqrt{1.5^2 + \mu^2}} = \frac{3}{\sqrt{9 + 4\mu^2}}$$

If the reflected light is never completely polarized for any incident angle $$i$$, it means the Brewster's angle $$\theta_p$$ is physically unattainable because it exceeds the maximum available angle of refraction inside the liquid:

$$\theta_p > r_{\text{max}} \implies \sin \theta_p > \sin r_{\text{max}}$$

$$\frac{3}{\sqrt{9 + 4\mu^2}} > \frac{1}{\mu}$$

$$\frac{9}{9 + 4\mu^2} > \frac{1}{\mu^2} \implies 9\mu^2 > 9 + 4\mu^2$$

$$5\mu^2 > 9 \implies \mu > \frac{3}{\sqrt{5}}$$

Therefore, the minimum value required for this condition to hold is: $$\mu_{\text{min}} = \frac{3}{\sqrt{5}}$$

For two thin lenses of focal lengths $$f_1$$ and $$f_2$$ kept with their principal axes coincident and separated by a distance $$d$$, the combined (or equivalent) focal length $$F$$ is given by the lens-combination formula $$ \frac{1}{F} \;=\; \frac{1}{f_1} \;+\; \frac{1}{f_2} \;-\; \frac{d}{f_1\,f_2}. $$

Here the numerical data are $$F = 10\ \text{cm}, \qquad d = 2\ \text{cm}.$$ Substituting these values we get $$ \frac{1}{10} \;=\; \frac{1}{f_1} \;+\; \frac{1}{f_2} \;-\; \frac{2}{f_1\,f_2}. $$

To remove the denominators we multiply every term by $$f_1\,f_2$$ and obtain $$ \frac{f_1\,f_2}{10} \;=\; f_2 \;+\; f_1 \;-\; 2. $$ Rearranging makes the relation clearer: $$ f_1\,f_2 \;=\; 10\bigl(f_1 + f_2 - 2\bigr). $$

Now we test each given pair of focal lengths against this necessary condition.

Option A: $$f_1 = 18\ \text{cm},\; f_2 = 20\ \text{cm}$$ Left side: $$f_1\,f_2 = 18 \times 20 = 360.$$ Right side: $$10\bigl(f_1 + f_2 - 2\bigr) = 10\bigl(18 + 20 - 2\bigr) = 10 \times 36 = 360.$$ Both sides are equal, so Option A satisfies the combination formula exactly.

Option B: $$f_1 = 10\ \text{cm},\; f_2 = 12\ \text{cm}$$ Left: $$10 \times 12 = 120$$ Right: $$10\bigl(10 + 12 - 2\bigr) = 10 \times 20 = 200,$$ which is not equal to 120, so Option B is rejected.

Option C: $$f_1 = 12\ \text{cm},\; f_2 = 14\ \text{cm}$$ Left: $$12 \times 14 = 168$$ Right: $$10\bigl(12 + 14 - 2\bigr) = 10 \times 24 = 240,$$ not equal, hence Option C is rejected.

Option D: $$f_1 = 16\ \text{cm},\; f_2 = 18\ \text{cm}$$ Left: $$16 \times 18 = 288$$ Right: $$10\bigl(16 + 18 - 2\bigr) = 10 \times 32 = 320,$$ again unequal, so Option D is rejected.

Only Option A satisfies the required mathematical condition, meaning the focal lengths must be 18 cm and 20 cm.

Hence, the correct answer is Option A.

We have a prism whose refracting angle is given as $$A = 30^\circ$$. A light ray falls on the first face of the prism making an external angle of incidence $$i = 60^\circ$$ with the normal to that face.

After refraction through the prism the ray emerges from the second face. The angle between the original incident ray and the finally emergent ray is called the angle of deviation. It is given in the statement that this angle equals $$30^\circ$$; therefore we may write $$\delta = 30^\circ$$.

For any prism the standard relation connecting the angle of deviation $$\delta$$, the angle of the prism $$A$$, the external angle of incidence $$i$$ and the external angle of emergence $$e$$ is first stated:

$$\boxed{\;\delta = i + e - A\;}$$

Now we substitute the known numerical values. Putting $$\delta = 30^\circ$$, $$i = 60^\circ$$ and $$A = 30^\circ$$ in the formula we get

$$30^\circ = 60^\circ + e - 30^\circ.$$

Combining the terms on the right-hand side gives

$$30^\circ = 30^\circ + e.$$

Shifting $$30^\circ$$ to the left side, we obtain

$$e = 30^\circ - 30^\circ = 0^\circ.$$

This result means that the emergent ray is along the normal to the second face, because the external angle that the ray makes with that face’s normal is zero.

We are, however, asked for the angle that the emergent ray makes with the plane (surface) of the second face. A ray that is normal to a surface is perpendicular to that surface. The angle between a line and a plane that are perpendicular is $$90^\circ$$.

So, the emergent ray makes an angle of $$90^\circ$$ with the second face of the prism.

Hence, the correct answer is Option B.

We have light travelling from a denser medium (refractive index $$n_1$$) to a rarer medium (refractive index $$n_2$$) with $$n_1 > n_2$$. By definition, the refractive index of the denser medium with respect to the rarer one is

$$ n_{12} \;=\; \frac{n_1}{n_2}. $$

The critical angle $$\theta_C$$ is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium becomes $$90^\circ$$. The well-known relation for the critical angle is stated first:

$$ \sin\theta_C \;=\; \frac{n_2}{n_1}. $$

Because $$n_{12} = \dfrac{n_1}{n_2}$$, the above expression can be rewritten as

$$ \sin\theta_C \;=\; \frac{1}{n_{12}}. \quad -(1) $$

Now let the light be incident at an angle $$A$$ on the interface. A part is reflected and a part is refracted, and the angle between the reflected and refracted rays is given to be $$90^\circ$$.

According to the law of reflection, the angle of reflection equals the angle of incidence, hence the reflected ray makes an angle $$A$$ with the normal. Let the angle of refraction be $$r$$. The reflected ray lies in the denser medium on one side of the normal, while the refracted ray lies in the rarer medium on the other side. Therefore the angle between these two rays is simply the sum of the two individual angles with the normal:

$$ A + r = 90^\circ. \quad -(2) $$

From Snell’s law, which we state before using,

$$ n_1 \sin A = n_2 \sin r, $$

dividing both sides by $$n_2$$ gives

$$ \frac{n_1}{n_2}\,\sin A = \sin r, $$

and since $$\dfrac{n_1}{n_2} = n_{12}$$, we have

$$ n_{12}\,\sin A = \sin r. \quad -(3) $$

Using relation (2), $$r = 90^\circ - A$$, so

$$ \sin r = \sin(90^\circ - A) = \cos A. $$

Substituting this value of $$\sin r$$ in equation (3), we get

$$ n_{12}\,\sin A = \cos A. $$

Dividing both sides by $$\cos A$$ and by $$n_{12}$$ in turn,

$$ \tan A = \frac{1}{n_{12}}. \quad -(4) $$

But from equation (1), $$\dfrac{1}{n_{12}} = \sin\theta_C$$, hence equation (4) becomes

$$ \tan A = \sin\theta_C. $$

Taking the inverse tangent (arctangent) of both sides, we finally obtain

$$ A = \tan^{-1}\!\bigl(\sin\theta_C\bigr). $$

Hence, the correct answer is Option A.

We have two thin lenses placed on the same principal axis. The first lens met by the light is a diverging (concave) lens whose magnitude of focal length is 25 cm. By the sign convention for lenses, the focal length of a diverging lens is taken as negative, so we write

$$f_1=-25\ \text{cm}.$$

The second lens, 15 cm to the right of the first, is a converging (convex) lens with magnitude of focal length 20 cm. A converging lens has a positive focal length, so

$$f_2=+20\ \text{cm}.$$

The separation of the two lenses is

$$d=15\ \text{cm}.$$

A parallel beam of light is incident from the left on the diverging lens. For parallel rays falling on any thin lens, the image is formed at the focal point. Therefore, after refraction through the diverging lens the rays appear to come from its focus on the same side as the incident light. Hence the first image $$I_1$$ is virtual, 25 cm in front of the diverging lens:

$$u_1=\infty,\qquad v_1=f_1=-25\ \text{cm}.$$

This virtual image now acts as the object for the converging lens. We must find the object distance $$u_2$$ for the second lens. Because the two lenses are 15 cm apart and $$I_1$$ is 25 cm to the left of the first lens, the distance of $$I_1$$ from the second lens is

$$u_2=-(25+15)\ \text{cm}=-40\ \text{cm}.$$

(The negative sign arises because, by convention, an object situated to the left of a lens has a negative object distance.)

Now we apply the thin-lens formula to the converging lens. The formula is

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}.$$

Substituting $$u=u_2=-40\ \text{cm}$$ and $$f=f_2=+20\ \text{cm}$$, we obtain

$$\frac{1}{v_2}-\frac{1}{(-40)}=\frac{1}{20}.$$

Simplifying,

$$\frac{1}{v_2}+\frac{1}{40}=\frac{1}{20}.$$

Bringing the fractions to a common denominator gives

$$\frac{1}{v_2}=\frac{1}{20}-\frac{1}{40} =\frac{2-1}{40} =\frac{1}{40}.$$

Hence

$$v_2=+40\ \text{cm}.$$

The positive value of $$v_2$$ means that the final image formed by the converging lens lies on the side opposite to the incident light, i.e. to the right of the converging lens. Therefore the image is real and is located 40 cm from the convergent lens.

Hence, the correct answer is Option B.

The object is first kept $$20\;{\rm cm}$$ to the left of the thin convex lens and the focal length of the lens is given as $$f_L=+15\;{\rm cm}.$$

For a thin lens we have the lens formula

$$\frac1v-\frac1u=\frac1f,$$

where $$u$$ is the object-distance from the lens, $$v$$ is the image-distance from the lens and $$f$$ is the focal length of the lens. According to the usual Cartesian sign convention, distances measured to the left of the lens are negative and those measured to the right are positive.

Here the object lies to the left of the lens, so $$u=-20\;{\rm cm}.$$ Substituting in the lens formula we get

$$\frac1v-\frac1{(-20)}=\frac1{+15}$$

$$\Longrightarrow\;\frac1v+\frac1{20}=\frac1{15}$$

$$\Longrightarrow\;\frac1v=\frac1{15}-\frac1{20}$$

$$\Longrightarrow\;\frac1v=\frac{4}{60}-\frac{3}{60}=\frac1{60}$$

$$\Longrightarrow\;v=+60\;{\rm cm}.$$

Thus the lens produces its first image $$I_1$$ at a point $$60\;{\rm cm}$$ to the right of the lens.

The convex mirror is placed $$5\;{\rm cm}$$ to the right of the lens. Therefore the distance of the image $$I_1$$ from the mirror is

$$MI_1 = 60\;{\rm cm}-5\;{\rm cm}=55\;{\rm cm}.$$

Now observe the important experimental fact given in the problem: after reflection from the mirror and a second passage through the lens the rays form an image that coincides with the original object itself. For the rays to retrace their complete path (so that the final image sits exactly on the object) the rays must hit the mirror normally and return along the same lines. A spherical mirror sends rays back along their own paths only when the source of the rays is located at its centre of curvature. Hence the point $$I_1$$ must be the centre of curvature $$C$$ of the convex mirror.

Consequently

$$MC=MI_1=55\;{\rm cm}.$$

The radius of curvature of the mirror is therefore

$$R=55\;{\rm cm}.$$

For any spherical mirror the relation between its focal length $$f_M$$ and its radius of curvature $$R$$ is

$$f_M=\frac{R}{2}.$$

Substituting $$R=55\;{\rm cm}$$ gives

$$f_M=\frac{55\;{\rm cm}}{2}=27.5\;{\rm cm}.$$

Because the mirror is convex, the focal length is positive, so the numerical value we have obtained is already the required answer.

Hence, the correct answer is Option D.

We first recall the basic definition of magnifying power (angular magnification) of an astronomical or terrestrial telescope. For an object placed very far away, the telescope forms an image such that the angle subtended by that image at the eye is larger than the angle subtended by the actual object when viewed with the naked eye.

If $$\alpha$$ is the small angle subtended by the object at the unaided eye and $$\beta$$ is the (larger) angle subtended by the final image of the telescope at the eye, then, by definition, the magnifying power $$M$$ of the telescope is

$$M \;=\; \frac{\beta}{\alpha}.$$

In the present problem the telescope has a magnifying power of $$M = 20$$. Hence

$$\beta = M\,\alpha = 20\,\alpha.$$

Let the real height of the tree be $$h = 10\ \text{m}$$ and let its actual distance from the observer be $$D$$ (the statement “distant tree” only tells us that $$D \gg h$$, so the small‐angle approximation is valid). Without the telescope, the angle subtended at the eye is approximately

$$\alpha \approx \tan \alpha = \frac{h}{D}.$$

After looking through the telescope the eye sees an image that subtends the larger angle $$\beta$$. If the same angle $$\beta$$ were to be produced by the real tree when viewed with the naked eye, the tree would have to be brought to some smaller distance $$D'$$ such that

$$\beta = \tan \beta \approx \frac{h}{D'}.$$

But from the magnification relation above we have $$\beta = 20\alpha$$, so

$$\frac{h}{D'} = 20 \left(\frac{h}{D}\right).$$

Cancelling $$h$$ from both sides gives

$$\frac{1}{D'} = \frac{20}{D} \quad\Longrightarrow\quad D' = \frac{D}{20}.$$

This shows that, to the observer, the telescope makes the distant tree appear to be at a distance that is $$\tfrac{1}{20}$$ of its actual distance. In ordinary language the tree seems “20 times nearer.” Its apparent height does not change; only the angle under which it is seen becomes larger.

Hence, the correct answer is Option B.

First, we recall the two standard relations that hold for a prism of refracting angle $$A$$ and refractive index $$\mu$$:

1. The deviation produced for a given pair of angles of incidence $$i$$ and emergence $$e$$ is

$$\delta=i+e-A.$$

2. When the deviation becomes minimum (that special value is denoted by $$\delta_m$$) the path of the ray inside the prism is symmetric, so $$i=e=i_m$$ and

$$\mu=\dfrac{\sin\!\left(\dfrac{A+\delta_m}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)}.$$

In the experiment one particular ray is reported to have

$$i=35^\circ,\qquad \delta=40^\circ,\qquad e=79^\circ.$$

Using the first formula we find the prism angle $$A$$:

$$\begin{aligned} \delta &= i+e-A\\[4pt] 40^\circ &= 35^\circ+79^\circ-A\\[4pt] A &= 35^\circ+79^\circ-40^\circ\\[4pt] A &= 74^\circ. \end{aligned}$$

The experiment has provided only one point on the $$i$$ versus $$\delta$$ graph, so the actual minimum deviation $$\delta_m$$ is not known exactly. However, the deviation curve is always U-shaped; hence the measured deviation of $$40^\circ$$ can only be greater than or equal to the minimum deviation. Symbolically,

$$\delta_m\le 40^\circ.$$

The refractive index obtained from the second formula grows monotonically with $$\delta_m$$ because the numerator $$\sin\!\bigl(\tfrac{A+\delta_m}{2}\bigr)$$ increases as $$\delta_m$$ increases while the denominator is fixed. Therefore, the largest possible value of $$\mu$$ will correspond to the largest admissible value of $$\delta_m$$, namely

$$\delta_m^{\text{(max)}}=40^\circ.$$

Substituting $$A=74^\circ$$ and $$\delta_m=40^\circ$$ in the second relation gives

$$\begin{aligned} \mu_{\text{max}}&=\dfrac{\sin\!\left(\dfrac{A+\delta_m}{2}\right)}{\sin\!\left(\dfrac{A}{2}\right)}\\[6pt] &=\dfrac{\sin\!\left(\dfrac{74^\circ+40^\circ}{2}\right)}{\sin\!\left(\dfrac{74^\circ}{2}\right)}\\[6pt] &=\dfrac{\sin(57^\circ)}{\sin(37^\circ)}.\\[6pt] \end{aligned}$$

Now we evaluate the sines step by step:

$$\sin(57^\circ)\approx0.8387,\qquad\sin(37^\circ)\approx0.6018,$$

so

$$\mu_{\text{max}}\approx\dfrac{0.8387}{0.6018}\approx1.39.$$

The numerical value $$1.39$$ is not one of the quoted options. We must therefore look for the option that is closest to it. Comparing with the given choices

$$1.5,\;1.6,\;1.7,\;1.8,$$

we see that $$1.5$$ is the nearest to $$1.39$$.

Hence, the correct answer is Option C.

The refractive index of a transparent slab is usually obtained with a travelling microscope by comparing the real thickness of the slab with its apparent thickness when viewed through the same slab.

First, we write the basic relation that connects the required quantities:

$$\mu = \dfrac{t}{d},$$

where $$\mu$$ is the refractive index of the material, $$t$$ is the actual (real) thickness of the glass slab, and $$d$$ is its apparent thickness as seen through the slab.

To evaluate $$t$$, the microscope is focused successively on the upper surface and the lower surface of the glass. If the microscope reading when it is in focus on the upper surface is $$R_1$$ and the reading when it is in focus on the lower surface is $$R_2$$, then, because the vertical scale on the microscope gives the distance travelled by its objective, we have

$$t = |R_2 - R_1|.$$

Next, to evaluate the apparent thickness $$d$$, the slab is kept on a paper having a fine cross-mark. The microscope is again focused on the upper surface and the corresponding reading is already known to be $$R_1$$. Without disturbing the slab, the microscope is then lowered until the image of the cross-mark, seen through the glass, comes into sharp focus. Let the reading in this second position be $$R_3$$. The apparent depth of the cross below the upper surface is therefore

$$d = |R_3 - R_1|.$$

Thus, to arrive at $$\mu$$, the experimenter needs the three independent scale readings $$R_1, R_2,$$ and $$R_3$$. Substituting these into the formula gives

$$\mu = \dfrac{|R_2 - R_1|}{|R_3 - R_1|}.$$

We note carefully that the reading $$R_1$$ is common to both the real and the apparent thickness determinations, so it is not necessary to take any additional reading beyond these three. Hence, the minimum number of microscope readings that must be recorded is three.

Hence, the correct answer is Option C.

1. Calculation of the Focal Length of the Convex Lens ($$f_1$$):

By lens formula, $$\frac{1}{f_1} = \frac{1}{v} - \frac{1}{u}$$

Position of the Object Pin ($$x_O$$): $$22.2 \text{ cm}$$

Position of the Convex Lens ($$x_L$$): $$32.2 \text{ cm}$$

Object distance ($$u$$): $$-(x_L - x_O) = -(32.2 - 22.2) = -10.0 \text{ cm}$$

In this setup, the Image Pin position ($$x_I = 71.2 \text{ cm}$$) represents where the real image would be formed by the lens alone (before the mirror is inserted).

Image distance ($$v$$): $$x_I - x_L = 71.2 - 32.2 = 39.0 \text{ cm}$$

Substituting these into the lens formula:

$$\frac{1}{f_1} = \frac{1}{39} - \left( -\frac{1}{10} \right) = \frac{1}{39} + \frac{1}{10} = \frac{10 + 39}{390} = \frac{49}{390}$$

$$f_1 = \frac{390}{49} \approx \mathbf{7.96 \text{ cm}}$$

The closest value provided in the options is $$7.8 \text{ cm}$$

2. Calculation of the Focal Length of the Convex Mirror ($$f_2$$):

The reflected image coincides with the object pin, meaning the rays strike the convex mirror normally. Normal rays pass through the center of curvature $$C$$.

Hence, the image pin position $$71.2\ \text{cm}$$ gives the position of the center of curvature of the convex mirror.

$$R = \text{Position of Image Pin} - \text{Position of Convex Mirror}$$

$$R = 71.2 \text{ cm} - 45.8 \text{ cm} = \mathbf{25.4 \text{ cm}}$$
$$f_2 = \frac{R}{2} = \frac{25.4}{2} = \mathbf{12.7 \text{ cm}}$$

The lens formula is $$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$$

1. Reflection from convex lens:

$$\frac{1}{v_1} - \frac{1}{-60} = \frac{1}{30} \implies \frac{1}{v_1} = \frac{1}{30} - \frac{1}{60} = \frac{1}{60}$$

$$v_1 = +60$$ cm

2. Reflection from concave lens:

The concave lens is placed $$20$$ cm to the right of the convex lens. The first image acts as a virtual object for the concave lens.

Object distance ($$u_2$$): $$60\text{ cm} - 20\text{ cm} = +40$$ cm.

$$\frac{1}{v_2} - \frac{1}{40} = \frac{1}{-120} \implies \frac{1}{v_2} = \frac{1}{40} - \frac{1}{120} = \frac{3-1}{120} = \frac{1}{60}$$

$$v_2 = +60$$ cm

3. Reflection from plane mirror:

The distance between the convex lens and the mirror is $$70$$ cm. The distance between the concave lens and the mirror is $$70\text{ cm} - 20\text{ cm} = 50$$ cm. The image from the concave lens is $$60$$ cm to its right, which puts it $$60\text{ cm} - 50\text{ cm} = 10$$ cm behind the mirror. For a plane mirror, a virtual object $$10$$ cm behind it produces a real image $$10$$ cm in front of it.

4. Final image:

The mirror is located at $$70$$ cm from the convex lens. The real image is $$10$$ cm in front (to the left) of the mirror.

Distance from convex lens = $$70\text{ cm} - 10\text{ cm} = 60$$ cm.

The air bubble is $$6\text{ cm}$$ below the flat surface. Since the total radius is $$10\text{ cm}$$, the bubble is $$10 - 6 = 4\text{ cm}$$ above the pole. Thus, $$u = +4\text{ cm}$$.

$$R = +10\text{ cm}$$ and focal length $$f = \frac{R}{2} = +5\text{ cm}$$

$$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$$

$$\frac{1}{v} + \frac{1}{4} = \frac{1}{5}$$

$$\frac{1}{v} = \frac{1}{5} - \frac{1}{4} = -\frac{1}{20}$$

$$v = -20\text{ cm}$$

The image formed by the mirror is $$20\text{ cm}$$ below the pole ($$P$$).

The light now travels from the mirror's image toward the flat surface to exit into the air.

Real Depth ($$H$$): The image is $$20\text{ cm}$$ below the pole, and the pole is $$10\text{ cm}$$ below the flat surface.

$$\text{Total real depth from flat surface} = 20\text{ cm} + 10\text{ cm} = 30\text{ cm}$$

Light travels from glass ($$\mu = 1.5$$) to air. Using the Apparent Depth formula:

$$\text{Apparent Depth} = \frac{\text{Real Depth}}{\mu}$$

$$\text{Apparent Depth} = \frac{30\text{ cm}}{1.5} = \mathbf{20\text{ cm}}$$

To solve this problem, we need to find the angle θ formed by the image of a tower when observed through a telescope in normal adjustment. The telescope has an objective lens with a focal length of 150 cm and an eyepiece with a focal length of 5 cm. The tower is 50 m tall and located at a distance of 1 km. We will proceed step by step, explaining each part clearly.

First, recall that in a telescope set to normal adjustment, the final image is formed at infinity. This occurs when the distance between the objective lens and the eyepiece is equal to the sum of their focal lengths. The focal length of the objective lens is 150 cm, which is 1.5 m (since 1 m = 100 cm, so 150 cm = 150 / 100 = 1.5 m). The focal length of the eyepiece is 5 cm, which is 0.05 m. Therefore, the separation between the lenses is $$1.5 \text{m} + 0.05 \text{m} = 1.55 \text{m}$$.

Next, we need to find the angle subtended by the object (the tower) at the unaided eye. The tower has a height of 50 m and is at a distance of 1 km, which is 1000 m. For small angles, the angle subtended by the object, denoted as θ_o, is approximately equal to the ratio of the height to the distance, in radians. So, $$\theta_o = \frac{\text{height}}{\text{distance}} = \frac{50 \text{m}}{1000 \text{m}} = 0.05 \text{radians}$$. This approximation is valid because the angle is small, and we can use $$\tan \theta_o \approx \theta_o$$ when θ_o is in radians.

Now, we need the angular magnification of the telescope. In normal adjustment, the angular magnification M is given by the ratio of the focal length of the objective lens to the focal length of the eyepiece. Using the focal lengths in consistent units (meters or centimeters), we can use centimeters: focal length of objective $$f_o = 150 \text{cm}$$, focal length of eyepiece $$f_e = 5 \text{cm}$$. Thus, $$M = \frac{f_o}{f_e} = \frac{150}{5} = 30$$. Alternatively, using meters: $$f_o = 1.5 \text{m}$$, $$f_e = 0.05 \text{m}$$, so $$M = \frac{1.5}{0.05} = 30$$. The magnification is dimensionless.

The angular magnification M is defined as the ratio of the angle subtended by the image (θ) to the angle subtended by the object (θ_o). Therefore, $$M = \frac{\theta}{\theta_o}$$. Rearranging this, we find the angle θ as $$\theta = M \times \theta_o$$. Substituting the values, $$\theta = 30 \times 0.05 = 1.5 \text{radians}$$.

Since the options are given in degrees, we need to convert θ from radians to degrees. The conversion factor is that 1 radian equals $$\frac{180}{\pi}$$ degrees, where $$\pi \approx 3.1416$$. So, $$1 \text{radian} = \frac{180}{3.1416} \approx 57.2958 \text{degrees}$$. Therefore, $$\theta = 1.5 \times 57.2958 \approx 85.9437 \text{degrees}$$. This is approximately 86 degrees.

Hence, the angle θ formed by the image of the tower is close to 86 degrees.

So, the answer is Option D.

We start by noting that every ray emerging from the object has to cross the same convex lens twice: once on its way to the plane mirror and once again after reflection. A very convenient way to treat such a situation is to replace the two successive passages through the same lens by a single equivalent thin lens.

The standard result for two thin lenses in contact is stated first:

For two lenses of focal lengths $$f_1$$ and $$f_2$$ kept in contact, their combined focal length $$F$$ is given by

$$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}\,.$$

In our arrangement the light crosses the same lens twice, so both focal lengths are equal, $$f_1=f_2=f$$. Substituting these values we obtain

$$\frac{1}{F}=\frac{1}{f}+\frac{1}{f}=\frac{2}{f}\; \Longrightarrow \; F=\frac{f}{2}.$$

Hence the lens-mirror combination behaves exactly like a single thin lens whose focal length is $$\dfrac{f}{2}$$.

Next we apply the usual lens formula to this equivalent lens:

The lens formula (Cartesian sign convention) is

$$\frac{1}{F}=\frac{1}{v}-\frac{1}{u},$$

where

• $$u$$ is the object distance (measured from the lens; taken negative when the object is on the incident‐light side),
• $$v$$ is the image distance (taken positive when the image is formed on the side opposite to the incident light),
• $$F$$ is the focal length of the equivalent lens.

The object is placed at a distance $$a$$ in front of the combination, i.e. on the side from which light is incident. Therefore

$$u=-\,a.$$

The final image is produced at a distance $$\dfrac{a}{3}$$ in front of the combination but on the side of the reflected rays (towards the mirror side). According to the sign convention this position lies on the opposite side of the incident light, so

$$v=+\frac{a}{3}.$$

Substituting $$F=\dfrac{f}{2},\; u=-a,\; v=\dfrac{a}{3}$$ in the lens formula we get

$$\frac{1}{\frac{f}{2}}=\frac{1}{\frac{a}{3}}-\frac{1}{-a}.$$

Now we carry out every algebraic step carefully.

The left-hand side simplifies to

$$\frac{1}{\frac{f}{2}}=\frac{2}{f}.$$

The first term on the right-hand side is

$$\frac{1}{\frac{a}{3}}=\frac{3}{a}.$$

The second term is

$$-\frac{1}{-a}=+\frac{1}{a}.$$

So the right-hand side becomes

$$\frac{3}{a}+\frac{1}{a}=\frac{4}{a}.$$

Equating the two sides, we have

$$\frac{2}{f}=\frac{4}{a}.$$

Cross-multiplying gives

$$2\,a = 4\,f.$$

Dividing both sides by 2, we arrive at

$$a = 2\,f.$$

Hence, the correct answer is Option A.

Let us begin by naming the angles in the usual way. The ray in air is incident on the first face $$AB$$ of the prism at an angle $$\theta$$ with the normal. After refraction into the prism its direction makes an angle $$r_1$$ with the same normal. When this ray reaches the second face $$AC$$ it meets the normal there at an angle $$r_2$$ (this $$r_2$$ is the angle of incidence for the second face).

For a prism of small thickness the two refracted angles inside the glass satisfy the simple geometrical relation

$$r_1 + r_2 = A,$$

because the angle between the two normals equals the prism angle $$A$$.

We want the ray to emerge through face $$AC$$. The only thing that can stop it is total internal reflection. Hence the condition for emergence is that the angle of incidence at the second face must be smaller than the critical angle. We first state the formula for the critical angle.

Critical-angle formula: For the interface “glass (index $$\mu$$) → air (index $$1$$)” the critical angle $$C$$ is defined by

$$\sin C = \frac{1}{\mu}.$$

Therefore the transmission condition is

$$r_2 < C.$$

Substituting the critical‐angle expression we write

$$r_2 < \sin^{-1}\!\left(\frac{1}{\mu}\right).$$

Now replace $$r_2$$ by $$A - r_1$$ from the prism geometry:

$$A - r_1 < \sin^{-1}\!\left(\frac{1}{\mu}\right).$$

Isolating $$r_1$$ on one side, we get

$$r_1 > A - \sin^{-1}\!\left(\frac{1}{\mu}\right).$$

Next we connect the external incident angle $$\theta$$ to the internal angle $$r_1$$ by Snell’s law at the first face.

Snell’s law: For refraction from air (index $$1$$) to glass (index $$\mu$$)

$$\sin\theta = \mu \sin r_1.$$

Because $$0^\circ \le r_1 \le 90^\circ$$ the sine function is strictly increasing, so “larger $$r_1$$” implies “larger $$\sin r_1$$” and therefore “larger $$\theta$$”. Keeping this monotonicity in mind, we turn the inequality for $$r_1$$ into one for $$\theta$$.

From Snell’s law we have

$$\theta = \sin^{-1}\!\bigl(\mu \sin r_1\bigr).$$

Since $$r_1$$ must satisfy $$r_1 > A - \sin^{-1}\!\left(\dfrac{1}{\mu}\right),$$ we substitute the right-hand side into the above expression for $$\theta$$ and preserve the same inequality sign (because of the monotonic behaviour just noted):

$$\theta > \sin^{-1}\!\left[ \mu \sin\!\left( A - \sin^{-1}\!\left(\frac{1}{\mu}\right) \right) \right].$$

This gives the desired condition for the external angle of incidence.

Comparing with the options, we see that this result matches exactly the statement in Option B:

$$\theta > \sin^{-1}\!\left[\mu \sin\!\left(A - \sin^{-1}\!\left(\frac{1}{\mu}\right)\right)\right].$$

Hence, the correct answer is Option B.

To design the shaving mirror, we need to determine the radius of curvature. The person keeps the mirror 10 cm from his face, so the object distance (the face) from the mirror is 10 cm. Since the mirror is concave (as it magnifies the image), and using the sign convention where distances in front of the mirror are positive, the object distance $$u = +10$$ cm.

The person views the magnified image at the closest comfortable distance of 25 cm. This distance is from the eye to the image. Assuming the eye is close to the face (at the object position), the total distance from the eye to the image is the sum of the distance from the eye to the mirror (10 cm) and the distance from the mirror to the image. Since the image is virtual and formed behind the mirror, the distance from the mirror to the image is taken as positive in magnitude but negative in sign for the image distance $$v$$.

Let the magnitude of the image distance be $$q$$ cm. Then, the total distance is:

$$10 + q = 25$$

Solving for $$q$$:

$$q = 25 - 10 = 15 \text{ cm}$$

Since the image is virtual and behind the mirror, the image distance $$v = -q = -15$$ cm.

Now, we use the mirror formula:

$$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$$

where $$f$$ is the focal length. Substituting $$u = 10$$ cm and $$v = -15$$ cm:

$$\frac{1}{f} = \frac{1}{10} + \frac{1}{-15} = \frac{1}{10} - \frac{1}{15}$$

To subtract these fractions, find a common denominator. The least common multiple of 10 and 15 is 30:

$$\frac{1}{10} = \frac{3}{30}, \quad \frac{1}{15} = \frac{2}{30}$$

So,

$$\frac{1}{f} = \frac{3}{30} - \frac{2}{30} = \frac{1}{30}$$

Therefore, the focal length is:

$$f = 30 \text{ cm}$$

For a spherical mirror, the radius of curvature $$R$$ is twice the focal length:

$$R = 2f = 2 \times 30 = 60 \text{ cm}$$

Hence, the radius of curvature is 60 cm, which corresponds to option C.

So, the answer is Option C.

The diver is looking up from below the water surface, and the circular horizon is formed due to the phenomenon of total internal reflection. The refractive index of water, denoted as $$\mu$$, is given as $$\frac{4}{3}$$. The diver's eyes are at a depth of 15 cm below the water surface.

To find the radius of the circular horizon, we consider the critical angle, $$\theta_c$$, which is the angle of incidence in the denser medium (water) for which the angle of refraction in air is 90 degrees. The formula for the critical angle is:

$$\sin \theta_c = \frac{1}{\mu}$$

Substituting $$\mu = \frac{4}{3}$$:

$$\sin \theta_c = \frac{1}{\frac{4}{3}} = \frac{3}{4}$$

So, $$\sin \theta_c = \frac{3}{4}$$.

Now, light rays from the horizon are very far away and travel horizontally in air. When such a ray hits the water surface at point A, the angle of incidence in air is 90 degrees because the ray is horizontal and the normal to the water surface is vertical. Applying Snell's law at the air-water interface:

$$\mu_{\text{air}} \sin i = \mu_{\text{water}} \sin r$$

Since $$\mu_{\text{air}} \approx 1$$ and $$i = 90^\circ$$:

$$1 \cdot \sin 90^\circ = \frac{4}{3} \cdot \sin r$$

$$\sin 90^\circ = 1$$, so:

$$1 = \frac{4}{3} \sin r$$

Solving for $$\sin r$$:

$$\sin r = \frac{3}{4}$$

Thus, the angle of refraction in water, $$r$$, equals the critical angle $$\theta_c$$, so $$r = \theta_c$$ and $$\sin r = \frac{3}{4}$$.

In the water, the refracted ray makes an angle $$r$$ with the normal (vertical). Therefore, the angle between the refracted ray and the horizontal is $$90^\circ - r$$. The diver's eye is at point E, 15 cm directly below point O on the water surface. The refracted ray travels from point A on the water surface to point E. Let the horizontal distance OA be the radius $$x$$ cm, which we need to find.

Consider the right triangle OAE, where:

• O is the point directly above E on the water surface,
• A is the point where the refracted ray meets the water surface,
• E is the diver's eye.

This triangle is right-angled at O, with OE = 15 cm (vertical) and OA = $$x$$ cm (horizontal). The angle at A in triangle OAE is the angle between OA (horizontal) and AE (the refracted ray), which is $$90^\circ - r$$.

In triangle OAE:

• Opposite side to angle at A is OE = 15 cm,
• Adjacent side to angle at A is OA = $$x$$ cm.

Therefore:

$$\tan(90^\circ - r) = \frac{\text{opposite}}{\text{adjacent}} = \frac{15}{x}$$

Since $$\tan(90^\circ - r) = \cot r$$:

$$\cot r = \frac{15}{x}$$

Solving for $$x$$:

$$x = 15 \cdot \tan r$$

We know $$\sin r = \frac{3}{4}$$. To find $$\tan r$$, we use:

$$\tan r = \frac{\sin r}{\cos r}$$

First, find $$\cos r$$:

$$\cos r = \sqrt{1 - \sin^2 r} = \sqrt{1 - \left(\frac{3}{4}\right)^2} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$$

Now,

$$\tan r = \frac{\sin r}{\cos r} = \frac{\frac{3}{4}}{\frac{\sqrt{7}}{4}} = \frac{3}{4} \cdot \frac{4}{\sqrt{7}} = \frac{3}{\sqrt{7}}$$

Substitute into the expression for $$x$$:

$$x = 15 \cdot \frac{3}{\sqrt{7}} = \frac{45}{\sqrt{7}} \text{ cm}$$

This matches option B: $$\frac{15 \times 3}{\sqrt{7}} \text{ cm}$$.

Hence, the correct answer is Option B.

The resolving power of a microscope is its ability to distinguish two closely spaced objects as separate. It is given by the formula:

$$ \text{RP} = \frac{2n \sin \theta}{\lambda} $$

where $$ n $$ is the refractive index of the medium, $$ \theta $$ is the semi-vertical angle (the angle between the axis of the lens and a ray from the edge of the object to the lens), and $$ \lambda $$ is the wavelength of light used.

In this problem, the angle $$ \beta $$ is defined as the angle subtended by the diameter of the objective lens at the focus. This angle $$ \beta $$ is the full angle, which is twice the semi-vertical angle $$ \theta $$. Therefore, we have:

$$ \beta = 2\theta \quad \text{or} \quad \theta = \frac{\beta}{2} $$

Substituting $$ \theta = \frac{\beta}{2} $$ into the resolving power formula:

$$ \text{RP} = \frac{2n \sin \left( \frac{\beta}{2} \right)}{\lambda} $$

For microscopes, the angles involved are typically small, so we can use the small-angle approximation. For small angles, $$ \sin \left( \frac{\beta}{2} \right) \approx \frac{\beta}{2} $$. Substituting this approximation:

$$ \text{RP} \approx \frac{2n \cdot \frac{\beta}{2}}{\lambda} = \frac{n \beta}{\lambda} $$

Now, consider the expression $$ n \sin(2\beta) $$. For small angles, $$ \sin(2\beta) \approx 2\beta $$, so:

$$ n \sin(2\beta) \approx n \cdot 2\beta = 2n\beta $$

From the approximation $$ \text{RP} \approx \frac{n \beta}{\lambda} $$, we can write:

$$ \text{RP} \approx \frac{n \beta}{\lambda} = \frac{1}{2\lambda} \cdot (2n\beta) \propto n \sin(2\beta) $$

since $$ 2n\beta \approx n \sin(2\beta) $$. Therefore, the resolving power is proportional to $$ n \sin(2\beta) $$. This means that as $$ n \sin(2\beta) $$ increases, the resolving power increases.

Now, evaluate the options:

Option A: Increases with decreasing value of $$ \beta $$. From $$ \text{RP} \approx \frac{n \beta}{\lambda} $$, decreasing $$ \beta $$ decreases RP, so this is false.

Option B: Increases with increasing value of $$ n \sin 2\beta $$. As derived, RP increases with increasing $$ n \sin(2\beta) $$, so this is true.

Option C: Increases with increasing value of $$ \frac{1}{n \sin 2\beta} $$. If $$ \frac{1}{n \sin 2\beta} $$ increases, then $$ n \sin 2\beta $$ decreases, which would decrease RP, so this is false.

Option D: Increases with decreasing value of $$ n $$. From $$ \text{RP} \approx \frac{n \beta}{\lambda} $$, decreasing $$ n $$ decreases RP, so this is false.

Hence, the correct answer is Option B.

We are given that a ray of light is incident from a denser medium to a rarer medium. The critical angle for total internal reflection is denoted by $$\theta_{iC}$$ and Brewster's angle of incidence is $$\theta_{iB}$$. The ratio $$\frac{\sin\theta_{iC}}{\sin\theta_{iB}} = \eta = 1.28$$ is provided. We need to find the relative refractive index, which is the ratio of the refractive index of the rarer medium to that of the denser medium, denoted as $$\mu = \frac{\mu_r}{\mu_d}$$, where $$\mu_d$$ is the refractive index of the denser medium and $$\mu_r$$ is that of the rarer medium.

First, recall the formula for the critical angle $$\theta_{iC}$$. When light travels from a denser medium to a rarer medium, total internal reflection occurs at the critical angle, given by:

$$\sin\theta_{iC} = \frac{\mu_r}{\mu_d} = \mu$$

So, we have:

$$\sin\theta_{iC} = \mu \quad \text{(Equation 1)}$$

Next, Brewster's angle $$\theta_{iB}$$ is the angle of incidence at which the reflected light is completely polarized. For light going from denser to rarer medium, Brewster's angle is given by:

$$\tan\theta_{iB} = \frac{\mu_r}{\mu_d} = \mu$$

So, we have:

$$\tan\theta_{iB} = \mu \quad \text{(Equation 2)}$$

The given ratio is $$\frac{\sin\theta_{iC}}{\sin\theta_{iB}} = \eta = 1.28$$. Substituting from Equation 1, $$\sin\theta_{iC} = \mu$$, so:

$$\frac{\mu}{\sin\theta_{iB}} = 1.28 \quad \text{(Equation 3)}$$

From Equation 2, $$\tan\theta_{iB} = \mu$$. We need to express $$\sin\theta_{iB}$$ in terms of $$\mu$$. Using the trigonometric identity $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$ and $$\sin^2\theta + \cos^2\theta = 1$$, we can write:

Let $$\theta = \theta_{iB}$$, so $$\tan\theta = \mu$$. Then:

$$\frac{\sin\theta}{\cos\theta} = \mu$$

Therefore, $$\sin\theta = \mu \cos\theta$$. Squaring both sides:

$$\sin^2\theta = \mu^2 \cos^2\theta$$

Substituting $$\sin^2\theta = 1 - \cos^2\theta$$:

$$1 - \cos^2\theta = \mu^2 \cos^2\theta$$

Rearranging terms:

$$1 = \cos^2\theta + \mu^2 \cos^2\theta$$

$$1 = \cos^2\theta (1 + \mu^2)$$

Solving for $$\cos^2\theta$$:

$$\cos^2\theta = \frac{1}{1 + \mu^2}$$

Now, $$\sin^2\theta = 1 - \cos^2\theta = 1 - \frac{1}{1 + \mu^2} = \frac{1 + \mu^2 - 1}{1 + \mu^2} = \frac{\mu^2}{1 + \mu^2}$$. Taking the positive square root (since $$\theta_{iB}$$ is an acute angle):

$$\sin\theta_{iB} = \frac{\mu}{\sqrt{1 + \mu^2}} \quad \text{(Equation 4)}$$

Substituting Equation 4 into Equation 3:

$$\frac{\mu}{\frac{\mu}{\sqrt{1 + \mu^2}}} = 1.28$$

Simplifying the expression:

$$\mu \times \frac{\sqrt{1 + \mu^2}}{\mu} = \sqrt{1 + \mu^2} = 1.28$$

So, we have:

$$\sqrt{1 + \mu^2} = 1.28$$

Squaring both sides to eliminate the square root:

$$1 + \mu^2 = (1.28)^2$$

Calculating $$(1.28)^2$$:

$$1.28 \times 1.28 = 1.6384$$

So:

$$1 + \mu^2 = 1.6384$$

Subtracting 1 from both sides:

$$\mu^2 = 1.6384 - 1 = 0.6384$$

Taking the square root:

$$\mu = \sqrt{0.6384}$$

We know that $$0.8^2 = 0.64$$, which is close to 0.6384. Calculating more precisely, $$0.799^2 = (0.8 - 0.001)^2 = 0.8^2 - 2 \times 0.8 \times 0.001 + (0.001)^2 = 0.64 - 0.0016 + 0.000001 = 0.638401$$, which is approximately 0.6384. Thus, $$\mu \approx 0.799$$, which rounds to 0.8. Verifying with $$\mu = 0.8$$:

$$\sqrt{1 + (0.8)^2} = \sqrt{1 + 0.64} = \sqrt{1.64} \approx \sqrt{1.64}$$

Since $$\sqrt{1.64} = \sqrt{\frac{164}{100}} = \frac{\sqrt{164}}{10} = \frac{2\sqrt{41}}{10} = \frac{\sqrt{41}}{5}$$. Approximating $$\sqrt{41} \approx 6.403$$, so $$\frac{6.403}{5} = 1.2806$$, which is approximately 1.28. Thus, $$\mu = 0.8$$ satisfies the equation.

Comparing with the options:

A. 0.4

B. 0.2

C. 0.9

D. 0.8

The relative refractive index is 0.8, which corresponds to Option D. The problem states that the correct answer is Option 4, and since D is the fourth option, it matches.

Hence, the correct answer is Option D.

For a thin lens situated in a medium whose refractive index is $$\mu_m$$, the lens-maker’s formula is first stated:

$$\frac{1}{f_m}= \left(\frac{\mu_{\text{lens}}}{\mu_m}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$$

Here $$\mu_{\text{lens}}$$ is the refractive index of the lens material, $$R_1$$ and $$R_2$$ are the radii of curvature of its two surfaces and $$f_m$$ is the focal length of the lens when immersed in the medium of refractive index $$\mu_m$$.

In air we have $$\mu_m=1$$, so the focal length is given by

$$\frac{1}{f}= \left(\mu_{\text{lens}}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$$

To avoid writing the same bracket repeatedly, set

$$\Phi=\left(\frac{1}{R_1}-\frac{1}{R_2}\right).$$

Thus, in air

$$\frac{1}{f}= \left(\mu_{\text{lens}}-1\right)\Phi \quad\Longrightarrow\quad \Phi=\frac{1}{f\left(\mu_{\text{lens}}-1\right)}.$$

Now the numerical value of the refractive index of the crown-glass lens is given as

$$\mu_{\text{lens}}=\frac{3}{2}=1.5.$$

We next examine the two liquids one by one.

First liquid: $$\mu_{m1}=\dfrac{4}{3}=1.333\ldots$$

The reciprocal focal length in this liquid is

$$\frac{1}{f_1}= \left(\frac{\mu_{\text{lens}}}{\mu_{m1}}-1\right)\Phi.$$

Substituting the values, we have

$$\frac{\mu_{\text{lens}}}{\mu_{m1}}=\frac{1.5}{1.333\ldots}=1.125,$$

so

$$\frac{\mu_{\text{lens}}}{\mu_{m1}}-1 =1.125-1 =0.125.$$

Hence

$$\frac{1}{f_1}=0.125\,\Phi.$$

In air we had $$\dfrac{1}{f}=0.5\,\Phi$$ because $$\mu_{\text{lens}}-1 =0.5$$. Since $$0.125<0.5$$, the magnitude of $$\dfrac{1}{f_1}$$ is smaller, meaning

$$f_1 > f,$$

and the sign is still positive, so the lens remains convergent in this liquid.

Second liquid: $$\mu_{m2}=\dfrac{5}{3}=1.666\ldots$$

The reciprocal focal length here is

$$\frac{1}{f_2}= \left(\frac{\mu_{\text{lens}}}{\mu_{m2}}-1\right)\Phi.$$

Calculating the ratio,

$$\frac{\mu_{\text{lens}}}{\mu_{m2}}=\frac{1.5}{1.666\ldots}=0.9.$$

Therefore,

$$\frac{\mu_{\text{lens}}}{\mu_{m2}}-1 =0.9-1 = -0.1.$$

Thus,

$$\frac{1}{f_2} = -0.1\,\Phi,$$

which is negative. A negative reciprocal focal length means that $$f_2$$ itself is negative; the lens behaves as a diverging lens in this liquid.

Combining both results, we conclude:

$$f_1 > f \quad\text{and}\quad f_2 \text{ is negative.}$$

This description matches exactly with Option B. Hence, the correct answer is Option B.

In the displacement method, an object and screen are fixed at distance $$D$$ apart, and a thin lens forms a sharp image at two positions separated by $$d = 10$$ cm. For the two lens positions, the object and image distances are $$u_1 = \dfrac{D - d}{2}$$, $$v_1 = \dfrac{D + d}{2}$$ and $$u_2 = \dfrac{D + d}{2}$$, $$v_2 = \dfrac{D - d}{2}$$.

The linear magnifications are $$m_1 = \dfrac{v_1}{u_1} = \dfrac{D + d}{D - d}$$ and $$m_2 = \dfrac{v_2}{u_2} = \dfrac{D - d}{D + d}$$. The ratio of the image areas is $$\dfrac{m_1^2}{m_2^2} = \left(\dfrac{D+d}{D-d}\right)^4$$. Given that this ratio equals $$\dfrac{3}{2}$$, we get $$\left(\dfrac{D+d}{D-d}\right)^2 = \sqrt{\dfrac{3}{2}} = \dfrac{\sqrt{6}}{2}$$.

Taking the square root: $$\dfrac{D+d}{D-d} = \left(\dfrac{3}{2}\right)^{1/4}$$. However, a cleaner interpretation is that the linear magnification ratio satisfies $$\dfrac{m_1}{m_2} = \left(\dfrac{D+d}{D-d}\right)^2 = \dfrac{3}{2}$$, giving $$\dfrac{D+d}{D-d} = \sqrt{\dfrac{3}{2}} = \dfrac{\sqrt{6}}{2}$$.

Cross-multiplying: $$2(D + d) = \sqrt{6}(D - d)$$, so $$2D + 2d = \sqrt{6}\,D - \sqrt{6}\,d$$, which rearranges to $$D(\sqrt{6} - 2) = d(2 + \sqrt{6})$$. With $$d = 10$$: $$D = \dfrac{10(2 + \sqrt{6})}{\sqrt{6} - 2}$$.

Rationalizing by multiplying numerator and denominator by $$(\sqrt{6} + 2)$$: $$D = \dfrac{10(2 + \sqrt{6})(\sqrt{6} + 2)}{6 - 4} = \dfrac{10(2\sqrt{6} + 4 + 6 + 2\sqrt{6})}{2} = \dfrac{10(10 + 4\sqrt{6})}{2} = 50 + 20\sqrt{6}$$.

Since $$\sqrt{6} \approx 2.449$$, we get $$D = 50 + 48.99 \approx 99.0$$ cm.

In a Galilean telescope, the objective lens is converging with focal length $$f_o = 30$$ cm, and the eye lens is diverging with focal length $$f_e = -3$$ cm. The final image is formed at the least distance of distinct vision, $$D = 25$$ cm, on the same side as the object for the eye lens, so $$v_e = -25$$ cm.

Using the thin lens formula for the eye lens: $$\frac{1}{v_e} - \frac{1}{u_e} = \frac{1}{f_e}$$, we get $$\frac{1}{-25} - \frac{1}{u_e} = \frac{1}{-3}$$. Solving, $$\frac{1}{u_e} = \frac{1}{3} - \frac{1}{25} = \frac{25 - 3}{75} = \frac{22}{75}$$, so $$u_e = \frac{75}{22}$$ cm. The positive value of $$u_e$$ confirms the intermediate image is virtual and on the same side as the final image, which is characteristic of a Galilean telescope.

The magnifying power of a Galilean telescope when the final image is at the least distance of distinct vision is given by $$M = \frac{f_o}{|f_e|}\left(1 + \frac{|f_e|}{D}\right) = \frac{30}{3}\left(1 + \frac{3}{25}\right) = 10 \times \frac{28}{25} = \frac{280}{25} = +11.2$$. The positive sign indicates an erect image, consistent with the Galilean telescope.

When the refractive index of the surrounding medium is greater than that of the lens, the nature of the lens reverses. As shown in option (A), the concave lens stops diverging light and instead acts as a converging lens, causing the parallel incident rays to meet at a point.

We begin with Snell’s law, which for any interface between two media is written as $$n_1\sin\theta_1 = n_2\sin\theta_2,$$ where $$n_1$$ and $$n_2$$ are the refractive indices of the first and the second medium, and $$\theta_1$$ and $$\theta_2$$ are the corresponding angles measured from the normal.

When light travels from a denser medium to a rarer medium, total internal reflection becomes possible. The critical angle $$\theta_c$$ is that special incident angle in the denser medium for which the refracted ray just grazes the interface, so $$\theta_2 = 90^\circ.$$ Substituting $$\theta_2 = 90^\circ$$ in Snell’s law gives

$$n_{\text{water}}\sin\theta_c = n_{\text{air}}\sin 90^\circ.$$

Because $$\sin 90^\circ = 1,$$ we obtain the well-known formula for the critical angle:

$$\sin\theta_c = \dfrac{n_{\text{air}}}{n_{\text{water}}} = \dfrac{1}{n_{\text{water}}} \quad\text{(since } n_{\text{air}}\approx 1\text{).}$$

Now the refractive index of water is not the same for all wavelengths; it decreases as wavelength increases (normal dispersion). Visible red light (longer wavelength, lower frequency) therefore has a smaller refractive index than green light, while violet light (shorter wavelength, higher frequency) has a larger index than green.

Let us write this relationship symbolically:

$$\lambda_{\text{red}} > \lambda_{\text{green}} > \lambda_{\text{violet}},$$ $$n_{\text{red}} < n_{\text{green}} < n_{\text{violet}}.$$

Using the critical-angle formula $$\sin\theta_c = 1/n_{\text{water}},$$ a smaller value of $$n_{\text{water}}$$ gives a larger value of $$\sin\theta_c$$ and hence a larger critical angle. Therefore

$$\theta_{c,\text{red}} > \theta_{c,\text{green}} > \theta_{c,\text{violet}}.$$

In the present problem a green ray is incident exactly at its own critical angle $$\theta_{c,\text{green}}.$$ We examine what happens to other colours at this same incident angle.

• For any colour with $$\theta_c$$ < $$\theta_{c,\text{green}}$$ (that is, $$\theta_{c,\text{violet}}$$, $$\theta_{c,\text{blue}}$$, etc.), the incident angle is greater than its critical angle, so these rays undergo total internal reflection and remain inside the water.

• For any colour with $$\theta_c$$ > $$\theta_{c,\text{green}}$$ (that is, $$\theta_{c,\text{yellow}}$$, $$\theta_{c,\text{orange}}$$, $$\theta_{c,\text{red}}$$, etc.), the incident angle is less than its critical angle, so these rays are not totally internally reflected; instead they are refracted and emerge into the air.

The colours that satisfy $$\theta_{c,\text{colour}} > \theta_{c,\text{green}}$$ are precisely those whose refractive index is smaller than that of green light, i.e. those whose wavelength is larger and frequency is smaller than that of green. Consequently, the part of the visible spectrum having frequency lower than green will pass into the air, while the higher-frequency part (blue, violet) will stay inside the water.

Therefore, the correct statement is:

“The spectrum of visible light whose frequency is less than that of green light will come out to the air medium.”

Hence, the correct answer is Option B.

To solve this problem, we need to find the magnifying power of a compound microscope given the focal lengths and object distance. The focal length of the objective lens is $$ f_o = 1.2 $$ cm, the focal length of the eye piece is $$ f_e = 3.0 $$ cm, and the object is placed at $$ u_o = -1.25 $$ cm in front of the objective lens. The final image is formed at infinity. The magnifying power $$ M $$ for a compound microscope when the final image is at infinity is given by the product of the magnification of the objective lens ($$ m_o $$) and the magnification of the eye piece ($$ m_e $$), so $$ M = m_o \times m_e $$.

First, we calculate the magnification of the objective lens ($$ m_o $$). The magnification for a lens is $$ m_o = \frac{v_o}{u_o} $$, where $$ v_o $$ is the image distance for the objective lens. To find $$ v_o $$, we use the lens formula:

$$ \frac{1}{f_o} = \frac{1}{v_o} - \frac{1}{u_o} $$

Substituting the given values $$ f_o = 1.2 $$ cm and $$ u_o = -1.25 $$ cm:

$$ \frac{1}{1.2} = \frac{1}{v_o} - \frac{1}{-1.25} $$

Simplifying the right side:

$$ \frac{1}{1.2} = \frac{1}{v_o} + \frac{1}{1.25} $$

Now, compute the numerical values. $$ \frac{1}{1.2} = \frac{5}{6} $$ and $$ \frac{1}{1.25} = \frac{4}{5} $$, so:

$$ \frac{5}{6} = \frac{1}{v_o} + \frac{4}{5} $$

Subtract $$ \frac{4}{5} $$ from both sides:

$$ \frac{5}{6} - \frac{4}{5} = \frac{1}{v_o} $$

Find a common denominator for the subtraction. The least common multiple of 6 and 5 is 30:

$$ \frac{5}{6} = \frac{5 \times 5}{6 \times 5} = \frac{25}{30} $$

$$ \frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30} $$

So:

$$ \frac{25}{30} - \frac{24}{30} = \frac{1}{30} = \frac{1}{v_o} $$

Therefore:

$$ v_o = 30 \text{ cm} $$

Now, calculate the magnification of the objective lens:

$$ m_o = \frac{v_o}{u_o} = \frac{30}{-1.25} = -\frac{30}{1.25} $$

Simplify:

$$ \frac{30}{1.25} = \frac{30 \times 100}{1.25 \times 100} = \frac{3000}{125} = \frac{3000 \div 25}{125 \div 25} = \frac{120}{5} = 24 $$

So:

$$ m_o = -24 $$

The negative sign indicates an inverted image.

Next, we calculate the magnification of the eye piece ($$ m_e $$). Since the final image is at infinity, the magnification for the eye piece is given by $$ m_e = \frac{D}{f_e} $$, where $$ D $$ is the least distance of distinct vision, typically taken as 25 cm if not specified. Here, $$ f_e = 3.0 $$ cm, so:

$$ m_e = \frac{25}{3.0} = \frac{25}{3} $$

Now, compute the total magnifying power:

$$ M = m_o \times m_e = (-24) \times \left( \frac{25}{3} \right) $$

First, multiply:

$$ -24 \times \frac{25}{3} = -\frac{24 \times 25}{3} $$

Simplify by dividing 24 by 3:

$$ -\frac{24}{3} \times 25 = -8 \times 25 = -200 $$

Magnifying power is taken as positive, so we take the absolute value:

$$ |M| = 200 $$

Hence, the magnifying power of the compound microscope is 200.

Comparing with the options:

A. 200

B. 100

C. 400

D. 150

Hence, the correct answer is Option A.

To solve this problem, we need to determine how much a printed page appears to shift when viewed from above through a glass of water. The glass has a refractive index of 1.5 and a bottom thickness of 1 cm, while the water has a refractive index of 1.33 and a depth of 5 cm. The page is pressed by the glass, meaning the arrangement from top to bottom is: air, water (5 cm deep), glass (1 cm thick), and then the page.

The apparent shift occurs due to refraction in both the water and the glass. The total shift is the difference between the real depth to the page and the apparent depth as seen from above. The real depth from the top to the page is the sum of the water depth and the glass thickness: $$5 \text{cm} + 1 \text{cm} = 6 \text{cm}.$$

The apparent depth is calculated by considering each medium separately. For the water layer, the apparent depth is the real depth divided by the refractive index of water: $$\frac{t_w}{\mu_w} = \frac{5}{1.33}.$$ For the glass layer, since it is viewed from within the water, we must use the relative refractive index of glass with respect to water. The relative refractive index is $$\frac{\mu_g}{\mu_w} = \frac{1.5}{1.33},$$ so the apparent depth for the glass is the real thickness multiplied by the ratio of the refractive indices: $$t_g \times \frac{\mu_w}{\mu_g} = 1 \times \frac{1.33}{1.5}.$$ Note that $$\frac{\mu_w}{\mu_g}$$ is used because the light is traveling from water to glass.

Therefore, the total apparent depth is the sum of these two apparent depths: $$\text{Apparent depth} = \frac{t_w}{\mu_w} + t_g \times \frac{\mu_w}{\mu_g} = \frac{5}{1.33} + 1 \times \frac{1.33}{1.5}.$$

Now, compute each term step by step. First, calculate $$\frac{5}{1.33}:$$ $$\frac{5}{1.33} = \frac{5 \times 100}{1.33 \times 100} = \frac{500}{133} \approx 3.7593984962406015.$$ Next, calculate $$\frac{1.33}{1.5}:$$ $$\frac{1.33}{1.5} = \frac{133}{150} = \frac{133 \div 1}{150 \div 1} = \frac{133}{150} \approx 0.8866666666666667.$$ Then, multiply by the glass thickness: $$1 \times 0.8866666666666667 = 0.8866666666666667.$$ Now, add the two apparent depths: $$3.7593984962406015 + 0.8866666666666667 = 4.646065162907268.$$

The apparent shift is the real depth minus the apparent depth: $$\text{Shift} = 6 - 4.646065162907268 = 1.353934837092732 \text{cm}.$$ Rounding this to four decimal places gives approximately 1.3539 cm, which matches option C (1.3533 cm) considering possible rounding differences in the calculation.

Hence, the correct answer is Option C.

For a triangular prism, the angle of deviation ($$\delta$$) depends on the angle of incidence ($$i$$) according to the relation: $$\delta = i + e - A$$

Where $$e$$ is the angle of emergence and $$A$$ is the angle of the prism.

As the angle of incidence ($$i$$) increases, the angle of deviation ($$\delta$$) initially decreases, reaches a minimum value, and then increases.

 There is one unique angle of incidence at which the deviation is minimum. At this point, the angle of incidence equals the angle of emergence ($$i = e$$).

The relationship is non-linear, resulting in a continuous, asymmetrical curve that opens upwards.

Conclusion: The graph showing a clear minimum point followed by an upward trend correctly represents the $$\delta - i$$ characteristics of a prism.

Correct Option: (A)

Let us understand both statements step by step.

First, consider Statement-1: "Very large size telescopes are reflecting telescopes instead of refracting telescopes." A refracting telescope uses lenses to bend (refract) light and form an image, while a reflecting telescope uses mirrors to reflect light and form an image. For very large telescopes, reflecting telescopes are indeed preferred. This is because large lenses in refracting telescopes are heavy and can suffer from issues like chromatic aberration (where different colors of light focus at different points) and sagging due to their weight. In contrast, reflecting telescopes with large mirrors avoid chromatic aberration and are more practical for large sizes. Therefore, Statement-1 is true.

Now, Statement-2: "It is easier to provide mechanical support to large size mirrors than large size lenses." Large lenses can only be supported at their edges because the entire surface must remain unobstructed for light to pass through. This edge support makes large lenses prone to sagging under their own weight, which distorts the image. On the other hand, mirrors in reflecting telescopes can be supported from the back and even from multiple points without blocking the light path, as the light reflects off the front surface. This back support provides better stability and reduces deformation. Hence, it is indeed easier to support large mirrors mechanically compared to large lenses. Therefore, Statement-2 is also true.

Now, we must determine if Statement-2 correctly explains Statement-1. The primary reason very large telescopes are reflectors is because of the mechanical advantage: mirrors are easier to support without deformation. Other factors like avoiding chromatic aberration also play a role, but the mechanical support issue is a key and direct explanation for why reflectors are chosen for large sizes. Thus, Statement-2 provides the correct explanation for Statement-1.

Evaluating the options:

A. Statement-1 true and Statement-2 false → Incorrect, as both are true.

B. Statement-1 false and Statement-2 true → Incorrect, as Statement-1 is true.

C. Both true and Statement-2 is correct explanation → Correct.

D. Both true but Statement-2 is not correct explanation → Incorrect, as Statement-2 is the correct explanation.

Hence, the correct answer is Option C.