Chapter 9: Ray Optics and Optical Instruments

Known data and theory

• A concave mirror of radius of curvature $$R$$ (focal length $$f=R/2$$).
• The mirror equation for any concave mirror is $$\dfrac1v+\dfrac1u=\dfrac1f$$, where $$u$$ is the object distance, $$v$$ is the image distance.
• For a given mirror, $$R$$ and hence $$f$$ depend only on the geometrical curvature, not on what fraction of the surface is exposed.
• Every small area element of a mirror produces its own complete image. The final image is the superposition of the images produced by all the illuminated elements.

Effect of covering the lower half

1. When the lower half is made opaque, the aperture (effective area) of the mirror is reduced to one-half.
   No change is produced in the radius of curvature $$R$$, so $$f$$ remains the same.
2. Hence, for the same object position $$u$$, the image position $$v$$ given by $$\dfrac1v+\dfrac1u=\dfrac1f$$ is unchanged.
3. Because the geometric rays that still reach the mirror (those striking the upper half) obey the same laws of reflection, they converge (or appear to diverge) from exactly the same point as before. Therefore
   • the location of the image,
   • its size (magnification $$m=-v/u$$), and
   • its nature (real or virtual, inverted or erect)
   remain exactly what they were for the uncovered mirror.
4. However, the number of rays contributing to the image is reduced approximately to half. Hence the luminous flux in the image falls to about 50 %, so the image becomes correspondingly less bright. (In practice the sharpness also drops slightly because of increased diffraction, but that is not the main point at this level.)

Ray diagram description

• Draw the principal axis, the pole P, focus F and centre of curvature C as usual.
• Show an object above the axis.
• Sketch only the upper half of the mirror surface as reflective; shade the lower half to indicate it is blocked.
• Draw two rays from the top of the object that strike the upper half: (i) one ray parallel to the principal axis, reflecting through F, (ii) a second ray passing through C (or directed towards C) and reflecting back on itself. Their intersection gives the top of the image at exactly the same place as for the full mirror.

Conclusion

Covering the lower half of the concave mirror does not affect the position, size or nature of the image; it only reduces its brightness.

Ray diagram to be drawn

• Draw the principal axis of a concave mirror M with its pole P, centre of curvature C and focus F marked.
• Place the mobile phone along the axis with its end A closer to the mirror and end B farther from it (A B is parallel to the axis).
• From A draw (i) a ray incident parallel to the axis which after reflection passes through F, and (ii) a ray through C which returns on itself. The two reflected rays meet at A′, the image of A.
• From B draw the same two standard rays. They intersect at B′, the image of B. Join A′B′ to obtain the complete image of the phone.

The construction shows that A′ and B′ are obtained at different image distances (v) from the mirror, and that the image is inverted.

Why the magnification is not uniform

For any point of the object at an axial distance u from the mirror, the mirror formula gives

$$\frac1v+\frac1u=\frac1f \;\;\Rightarrow\;\; v = \frac{fu}{u-f}$$

The linear magnification for that point is

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

Because the phone is an extended object lying along the axis, the nearer end A has a smaller object distance $$u_A$$ than the far end B (object distance $$u_B$$). Hence

$$m_A = -\frac{f}{u_A-f}\;\;\neq\;\;m_B = -\frac{f}{u_B-f}$$

Each small element of the phone is therefore magnified by a different factor; the image is not stretched or shrunk uniformly and consequently looks distorted (e.g. end A appears larger than end B).

Dependence of distortion on the position of the phone

• If the whole phone is far from the mirror (all $$u \gg f$$) or is close to the centre of curvature ($$u \approx 2f$$), then $$m = -f/(u-f)$$ varies very slowly with u; the difference between $$m_A$$ and $$m_B$$ is small and the distortion is negligible.
• If any point of the phone is near the focal point ($$u \rightarrow f$$), the denominator $$u-f$$ becomes very small, making $$m$$ extremely large for that part while the other end still has a moderate magnification. The image is then highly distorted.

Hence, the degree of distortion does depend on where the phone is placed with respect to the mirror; the closer any portion of it is to the focal point, the more pronounced is the distortion.

The mirror is concave, so its focal length is taken negative. The radius of curvature is $$R = 15 \, \mathrm{cm}$$, hence $$f = \dfrac{R}{2} = \dfrac{15}{2} = 7.5 \, \mathrm{cm}$$, and with the sign convention $$f = -7.5 \, \mathrm{cm}$$.

Case (i): object at 10 cm. Here $$u = -10 \, \mathrm{cm}$$. The mirror equation $$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$$ gives

$$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{-7.5} - \dfrac{1}{-10} = -\dfrac{1}{7.5} + \dfrac{1}{10} = \dfrac{-4 + 3}{30} = -\dfrac{1}{30}$$

$$\implies v = -30 \, \mathrm{cm}$$

The image forms $$30 \, \mathrm{cm}$$ in front of the mirror; the negative sign shows it is real and on the same side as the object.

Magnification: $$m = -\dfrac{v}{u} = -\dfrac{-30}{-10} = -3$$. The negative value means the image is inverted, and $$|m| = 3$$ means it is magnified three times.

Case (ii): object at 5 cm. Here $$u = -5 \, \mathrm{cm}$$, which lies between the pole and the focus. Then

$$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = -\dfrac{1}{7.5} + \dfrac{1}{5} = \dfrac{-2 + 3}{15} = \dfrac{1}{15}$$

$$\implies v = +15 \, \mathrm{cm}$$

The positive sign shows the image is $$15 \, \mathrm{cm}$$ behind the mirror, so it is virtual and erect.

Magnification: $$m = -\dfrac{v}{u} = -\dfrac{15}{-5} = +3$$. The image is erect and magnified three times.

The side-view mirror is convex, so its focal length is positive: $$f = \dfrac{R}{2} = \dfrac{2}{2} = +1 \, \mathrm{m}$$.

From the mirror equation $$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$$, the image distance is $$v = \dfrac{uf}{u - f}$$.

The jogger runs at a steady $$5 \, \mathrm{m\,s^{-1}}$$, so in each $$1 \, \mathrm{s}$$ interval the object distance changes by $$5 \, \mathrm{m}$$. For every position we compute the image location at the start of that second and one second later; the shift, divided by $$1 \, \mathrm{s}$$, is the speed of the image.

(a) Jogger 39 m away ($$u$$ from $$-39 \, \mathrm{m}$$ to $$-34 \, \mathrm{m}$$):

$$v_1 = \dfrac{(-39)(1)}{-39 - 1} = \dfrac{39}{40} \, \mathrm{m}, \qquad v_2 = \dfrac{(-34)(1)}{-34 - 1} = \dfrac{34}{35} \, \mathrm{m}$$

Shift $$= \dfrac{39}{40} - \dfrac{34}{35} = \dfrac{273 - 272}{280} = \dfrac{1}{280} \, \mathrm{m}$$, so the image speed is $$\dfrac{1}{280} \, \mathrm{m\,s^{-1}} \approx 3.6 \times 10^{-3} \, \mathrm{m\,s^{-1}}$$.

(b) Jogger 29 m away ($$u$$ from $$-29 \, \mathrm{m}$$ to $$-24 \, \mathrm{m}$$):

$$v_1 = \dfrac{29}{30} \, \mathrm{m}, \qquad v_2 = \dfrac{24}{25} \, \mathrm{m}$$

Shift $$= \dfrac{29}{30} - \dfrac{24}{25} = \dfrac{145 - 144}{150} = \dfrac{1}{150} \, \mathrm{m}$$, so the image speed is $$\dfrac{1}{150} \, \mathrm{m\,s^{-1}}$$.

(c) Jogger 19 m away ($$u$$ from $$-19 \, \mathrm{m}$$ to $$-14 \, \mathrm{m}$$):

$$v_1 = \dfrac{19}{20} \, \mathrm{m}, \qquad v_2 = \dfrac{14}{15} \, \mathrm{m}$$

Shift $$= \dfrac{19}{20} - \dfrac{14}{15} = \dfrac{57 - 56}{60} = \dfrac{1}{60} \, \mathrm{m}$$, so the image speed is $$\dfrac{1}{60} \, \mathrm{m\,s^{-1}}$$.

(d) Jogger 9 m away ($$u$$ from $$-9 \, \mathrm{m}$$ to $$-4 \, \mathrm{m}$$):

$$v_1 = \dfrac{9}{10} \, \mathrm{m}, \qquad v_2 = \dfrac{4}{5} \, \mathrm{m}$$

Shift $$= \dfrac{9}{10} - \dfrac{4}{5} = \dfrac{9 - 8}{10} = \dfrac{1}{10} \, \mathrm{m}$$, so the image speed is $$\dfrac{1}{10} \, \mathrm{m\,s^{-1}}$$.

Although the jogger moves at a constant $$5 \, \mathrm{m\,s^{-1}}$$, the image moves faster and faster as the jogger comes nearer. This is why a vehicle far behind, seen in a convex side-mirror, seems almost stationary, while a nearby one appears to rush towards you.

Light passes from air into glass through a single spherical surface. Apply $$\dfrac{n_2}{v} - \dfrac{n_1}{u} = \dfrac{n_2 - n_1}{R}$$ with $$n_1 = 1$$ (air) and $$n_2 = 1.5$$ (glass).

The surface is convex towards the incident light, so $$R = +20 \, \mathrm{cm}$$. The point source is a real object at $$u = -100 \, \mathrm{cm}$$.

$$\dfrac{1.5}{v} - \dfrac{1}{-100} = \dfrac{1.5 - 1}{20}$$

$$\dfrac{1.5}{v} + \dfrac{1}{100} = \dfrac{0.5}{20} = \dfrac{1}{40}$$

$$\dfrac{1.5}{v} = \dfrac{1}{40} - \dfrac{1}{100} = \dfrac{5 - 2}{200} = \dfrac{3}{200}$$

$$v = \dfrac{1.5 \times 200}{3} = 100 \, \mathrm{cm}$$

The image is formed $$100 \, \mathrm{cm}$$ from the spherical surface, on the glass side, in the direction of the incident light. The positive value of $$v$$ shows the image is real.

We see a lens because light bends as it crosses the boundary between the lens and its surroundings. When a lens of refractive index $$n_{\text{lens}}$$ is placed in a medium of refractive index $$n_m$$, its focal length is given by

$$\dfrac{1}{f} = \left( \dfrac{n_{\text{lens}}}{n_m} - 1 \right)\left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right)$$

If the liquid has the same refractive index as the glass, then $$n_m = n_{\text{lens}} = 1.47$$, so $$\dfrac{n_{\text{lens}}}{n_m} = 1$$ and

$$\dfrac{1}{f} = (1 - 1)\left( \dfrac{1}{R_1} - \dfrac{1}{R_2} \right) = 0 \implies f \to \infty$$

With an infinite focal length the lens has no converging or diverging power; it does not bend light at all and so becomes invisible — it appears to disappear.

Hence the refractive index of the liquid must be $$1.47$$. Water has refractive index $$1.33 \neq 1.47$$, so the liquid cannot be water.

(i) The power of a lens is $$P = \dfrac{1}{f}$$ with $$f$$ in metres. Here $$f = 0.5 \, \mathrm{m}$$, so

$$P = \dfrac{1}{0.5} = +2 \, \mathrm{D}$$

The positive sign shows it is a converging (convex) lens.

(ii) For a double convex lens the first face is convex to the incident light ($$R_1 = +10 \, \mathrm{cm}$$) and the second is concave to it ($$R_2 = -15 \, \mathrm{cm}$$). The lens maker's formula gives

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

$$\dfrac{1}{12} = (n - 1)\left( \dfrac{1}{10} - \dfrac{1}{-15} \right) = (n - 1)\left( \dfrac{1}{10} + \dfrac{1}{15} \right) = (n - 1)\dfrac{3 + 2}{30} = \dfrac{n - 1}{6}$$

$$n - 1 = \dfrac{6}{12} = 0.5 \implies n = 1.5$$

(iii) Let $$X = \dfrac{1}{R_1} - \dfrac{1}{R_2}$$ be the fixed geometric factor of the lens. In air,

$$\dfrac{1}{f_{\text{air}}} = (n_g - 1)\,X \implies \dfrac{1}{20} = (1.5 - 1)\,X \implies X = 0.1 \, \mathrm{cm^{-1}}$$

In water the lens material has refractive index $$\dfrac{n_g}{n_w} = \dfrac{1.5}{1.33}$$ relative to its surroundings, so

$$\dfrac{1}{f_{\text{water}}} = \left( \dfrac{n_g}{n_w} - 1 \right) X = \left( \dfrac{1.5}{1.33} - 1 \right)(0.1)$$

$$\dfrac{1}{f_{\text{water}}} = (1.128 - 1)(0.1) = (0.128)(0.1) = 0.0128 \, \mathrm{cm^{-1}}$$

$$f_{\text{water}} = \dfrac{1}{0.0128} \approx 78.2 \, \mathrm{cm}$$

The focal length in water is nearly four times that in air: the lens is much weaker because glass and water differ far less in refractive index than glass and air.

The complete step–by–step solution cannot be written because the numerical data that appear in Fig. 9.20 of the NCERT textbook are not available to the model. Without the focal-lengths of the two lenses, the object distance and the separation between the lenses, no unambiguous calculation is possible.

To obtain a rigorous worked answer, first note down from your copy of Fig. 9.20:

• focal length $$f_1$$ of lens L₁
• focal length $$f_2$$ of lens L₂
• separation $$d$$ between the two lenses
• object distance $$u_1$$ from L₁

Then carry out the following universally valid steps.

1. For L₁ use the thin-lens formula
   $$\frac{1}{f_1}=\frac{1}{v_1}-\frac{1}{u_1}$$
   to get the image distance $$v_1$$ from L₁.
2. Convert this image into the object for L₂. The object distance for the second lens is
   $$u_2= d-v_1$$
   taking the usual Cartesian sign convention (a positive value means the point lies to the right of L₂, a negative value to the left).
3. Apply the thin-lens formula to L₂
   $$\frac{1}{f_2}=\frac{1}{v_2}-\frac{1}{u_2}$$
   to get the final image distance $$v_2$$ measured from L₂.
4. The position of the overall image, measured from L₁, is therefore
   $$x = d+v_2.$$

Insert the four numerical values from your figure into these four equations and work through the algebra exactly as shown; every Class 12 Physics student should be able to do so once the actual numbers are to hand.

Once you have substituted, solved and kept signs consistent, you will obtain a single numerical answer for $$v_2$$ (and hence for $$x$$) that completely specifies the image location required in Example 9.8.

The mirror is concave, so $$f = \dfrac{R}{2} = \dfrac{36}{2} = 18 \, \mathrm{cm}$$, and with the sign convention $$f = -18 \, \mathrm{cm}$$. The candle is a real object: $$u = -27 \, \mathrm{cm}$$; its size is $$h = 2.5 \, \mathrm{cm}$$.

Mirror equation: $$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$$

$$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{-18} - \dfrac{1}{-27} = -\dfrac{1}{18} + \dfrac{1}{27} = \dfrac{-3 + 2}{54} = -\dfrac{1}{54}$$

$$\implies v = -54 \, \mathrm{cm}$$

The screen must be placed $$54 \, \mathrm{cm}$$ in front of the mirror (on the same side as the candle) to catch a sharp image. The negative sign shows the image is real.

Magnification: $$m = -\dfrac{v}{u} = -\dfrac{-54}{-27} = -2$$

The magnification is negative, so the image is inverted; its magnitude $$|m| = 2$$ means the image is enlarged two times. The height of the image is

$$h' = |m|\,h = 2 \times 2.5 = 5 \, \mathrm{cm}$$

So the screen receives a real, inverted image $$5 \, \mathrm{cm}$$ tall.

Moving the candle closer: as the candle approaches the focus ($$18 \, \mathrm{cm}$$), the image distance grows, so the screen must be moved farther away from the mirror. When the candle reaches the focus, the reflected rays become parallel and no image can be formed on a screen; if the candle is brought still nearer (between focus and pole) the image becomes virtual and cannot be caught on a screen at all.

The mirror is convex, so $$f = +15 \, \mathrm{cm}$$. The needle is a real object: $$u = -12 \, \mathrm{cm}$$; its size is $$h = 4.5 \, \mathrm{cm}$$.

Mirror equation: $$\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$$

$$\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} = \dfrac{1}{15} - \dfrac{1}{-12} = \dfrac{1}{15} + \dfrac{1}{12} = \dfrac{4 + 5}{60} = \dfrac{9}{60} = \dfrac{3}{20}$$

$$\implies v = \dfrac{20}{3} \approx 6.7 \, \mathrm{cm}$$

The positive value shows the image is formed about $$6.7 \, \mathrm{cm}$$ behind the mirror; it is virtual and erect.

Magnification: $$m = -\dfrac{v}{u} = -\dfrac{20/3}{-12} = \dfrac{20}{36} = \dfrac{5}{9} \approx 0.56$$

Image size $$= |m|\,h = \dfrac{5}{9} \times 4.5 = 2.5 \, \mathrm{cm}$$ — diminished and erect.

Moving the needle farther away: as $$u \to \infty$$, the image distance $$v \to f = 15 \, \mathrm{cm}$$, so the image moves away from the pole towards the focus, while the magnification steadily decreases — the image keeps getting smaller, always remaining virtual and erect.

Refractive index of water. For viewing nearly along the normal, $$n = \dfrac{\text{real depth}}{\text{apparent depth}}$$.

$$n = \dfrac{12.5}{9.4} \approx 1.33$$

Replacing water by the liquid ($$n' = 1.63$$). The real depth is unchanged at $$12.5 \, \mathrm{cm}$$, so the new apparent depth is

$$d' = \dfrac{\text{real depth}}{n'} = \dfrac{12.5}{1.63} \approx 7.67 \, \mathrm{cm}$$

Earlier the needle appeared to lie at a depth of $$9.4 \, \mathrm{cm}$$; now it appears at $$7.67 \, \mathrm{cm}$$. Since the apparent position has risen towards the surface, the microscope must be moved upward by

$$9.4 - 7.67 = 1.73 \, \mathrm{cm}$$

Let

• $$i_1=60^{\circ}$$ be the angle of incidence in air in Fig. 9.27(a).
• $$r_1=35^{\circ}$$ be the corresponding angle of refraction in glass as read from the figure.
• $$i_2=60^{\circ}$$ be the angle of incidence in air in Fig. 9.27(b).
• $$r_2=40^{\circ}$$ be the corresponding angle of refraction in water as read from the figure.

1. Refractive index of glass

Snell’s law for Fig. 9.27(a) (air → glass)

$$n_g=\frac{\sin i_1}{\sin r_1}=\frac{\sin 60^{\circ}}{\sin 35^{\circ}}\approx\frac{0.866}{0.574}=1.51$$

2. Refractive index of water

Snell’s law for Fig. 9.27(b) (air → water)

$$n_w=\frac{\sin i_2}{\sin r_2}=\frac{\sin 60^{\circ}}{\sin 40^{\circ}}\approx\frac{0.866}{0.643}=1.35$$

3. Refraction at the water–glass interface

In Fig. 9.27(c) a ray is incident from water onto glass with

$$i_3=45^{\circ}, \qquad r_3=\text{?}$$

Applying Snell’s law (water → glass)

$$n_w\,\sin i_3 = n_g\,\sin r_3$$

$$\Rightarrow\;\sin r_3 = \frac{n_w}{n_g}\,\sin 45^{\circ}$$

Substituting $$n_w$$ and $$n_g$$ obtained above,

$$\sin r_3 = \frac{1.35}{1.51}\,(0.707) = 0.632$$

$$r_3 = \sin^{-1}(0.632) \approx 39^{\circ}$$

Therefore, the ray is refracted in glass at approximately $$39^{\circ}$$ with the normal.

A ray from the bulb can leave the water only if it strikes the surface at an angle (to the normal) less than the critical angle $$C$$. A ray meeting the surface at exactly $$C$$ grazes along it; any ray beyond $$C$$ is totally internally reflected. So light escapes only through a circular patch of water directly above the bulb.

Critical angle. $$\sin C = \dfrac{1}{n} = \dfrac{1}{1.33} = 0.752 \implies C \approx 48.8^\circ$$

Radius of the circular patch. With the bulb at depth $$d = 80 \, \mathrm{cm}$$, the limiting ray makes angle $$C$$ with the vertical, so the radius of the patch is $$r = d\tan C$$.

$$\cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - (0.752)^2} = \sqrt{0.4347} = 0.659$$

$$\tan C = \dfrac{\sin C}{\cos C} = \dfrac{0.752}{0.659} = 1.14$$

$$r = 80 \times 1.14 = 91.2 \, \mathrm{cm} = 0.912 \, \mathrm{m}$$

Area of the patch.

$$A = \pi r^2 = \pi (0.912)^2 = \pi (0.8317) \approx 2.61 \, \mathrm{m^2}$$

Refractive index of the prism. For a prism of refracting angle $$A$$ at minimum deviation $$D_m$$,

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

With $$A = 60^\circ$$ and $$D_m = 40^\circ$$:

$$n = \dfrac{\sin\!\left( \dfrac{60^\circ + 40^\circ}{2} \right)}{\sin\!\left( \dfrac{60^\circ}{2} \right)} = \dfrac{\sin 50^\circ}{\sin 30^\circ} = \dfrac{0.766}{0.5} = 1.532$$

Prism placed in water. Now the relevant index is that of the glass relative to water:

$$n' = \dfrac{n_{\text{glass}}}{n_{\text{water}}} = \dfrac{1.532}{1.33} = 1.152$$

Using the same formula with the new minimum deviation $$D_m'$$:

$$\sin\!\left( \dfrac{A + D_m'}{2} \right) = n' \sin\!\left( \dfrac{A}{2} \right) = 1.152 \times \sin 30^\circ = 1.152 \times 0.5 = 0.576$$

$$\dfrac{A + D_m'}{2} = \sin^{-1}(0.576) \approx 35.2^\circ$$

$$A + D_m' = 70.3^\circ \implies D_m' = 70.3^\circ - 60^\circ \approx 10.3^\circ$$

For a double convex lens with both faces of the same radius $$R$$, the first face is convex to the incident light ($$R_1 = +R$$) and the second is concave to it ($$R_2 = -R$$). The lens maker's formula gives

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

With $$n = 1.55$$ and $$f = 20 \, \mathrm{cm}$$:

$$\dfrac{1}{20} = (1.55 - 1)\dfrac{2}{R} = 0.55 \times \dfrac{2}{R} = \dfrac{1.1}{R}$$

$$R = 1.1 \times 20 = 22 \, \mathrm{cm}$$

Without the lens the beam would converge to the point P. When the lens is interposed before the rays reach P, the point P acts as a virtual object for the lens. A virtual object lies on the far side, in the direction the light is travelling, so its distance is taken positive: $$u = +12 \, \mathrm{cm}$$.

The lens formula is $$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$$.

(a) Convex lens, $$f = +20 \, \mathrm{cm}$$:

$$\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{20} + \dfrac{1}{12} = \dfrac{3 + 5}{60} = \dfrac{8}{60} = \dfrac{2}{15}$$

$$v = \dfrac{15}{2} = 7.5 \, \mathrm{cm}$$

The beam now converges $$7.5 \, \mathrm{cm}$$ from the lens, on the side away from the lens. The convex lens makes the already-converging beam converge sooner.

(b) Concave lens, $$f = -16 \, \mathrm{cm}$$:

$$\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{-16} + \dfrac{1}{12} = \dfrac{-3 + 4}{48} = \dfrac{1}{48}$$

$$v = 48 \, \mathrm{cm}$$

The beam now converges $$48 \, \mathrm{cm}$$ from the lens. The concave lens diverges the beam, so it converges much farther away than P would have been.

The lens is concave, so $$f = -21 \, \mathrm{cm}$$. The object is real: $$u = -14 \, \mathrm{cm}$$; its size is $$h = 3.0 \, \mathrm{cm}$$.

Lens formula: $$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$$

$$\dfrac{1}{v} = \dfrac{1}{f} + \dfrac{1}{u} = \dfrac{1}{-21} + \dfrac{1}{-14} = -\dfrac{1}{21} - \dfrac{1}{14} = \dfrac{-2 - 3}{42} = -\dfrac{5}{42}$$

$$\implies v = -\dfrac{42}{5} = -8.4 \, \mathrm{cm}$$

The image is formed $$8.4 \, \mathrm{cm}$$ from the lens, on the same side as the object — so it is virtual and erect.

Magnification: $$m = \dfrac{v}{u} = \dfrac{-8.4}{-14} = +0.6$$

Image size $$= m\,h = 0.6 \times 3.0 = 1.8 \, \mathrm{cm}$$ — diminished.

Moving the object farther away: as $$u \to \infty$$, $$v \to f = -21 \, \mathrm{cm}$$, so the image moves away from the lens towards the focus (but never beyond it) and becomes smaller and smaller. A concave lens always forms a virtual, erect, diminished image, located between the lens and its focus.

For two thin lenses placed in contact, the powers add, so the combined focal length $$F$$ satisfies

$$\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2}$$

The convex lens has $$f_1 = +30 \, \mathrm{cm}$$ and the concave lens has $$f_2 = -20 \, \mathrm{cm}$$:

$$\dfrac{1}{F} = \dfrac{1}{30} + \dfrac{1}{-20} = \dfrac{1}{30} - \dfrac{1}{20} = \dfrac{2 - 3}{60} = -\dfrac{1}{60}$$

$$F = -60 \, \mathrm{cm}$$

The negative focal length shows the combination behaves as a diverging lens of focal length $$60 \, \mathrm{cm}$$.

Objective focal length $$f_o = 2.0 \, \mathrm{cm}$$, eyepiece focal length $$f_e = 6.25 \, \mathrm{cm}$$, separation between the lenses $$L = 15 \, \mathrm{cm}$$. The method is to work from the eyepiece backwards: first locate where the intermediate image must lie, then find the object position for the objective.

(a) Final image at the near point, $$D = 25 \, \mathrm{cm}$$.

For the eyepiece the final (virtual) image is at $$v_e = -25 \, \mathrm{cm}$$:

$$\dfrac{1}{u_e} = \dfrac{1}{v_e} - \dfrac{1}{f_e} = \dfrac{1}{-25} - \dfrac{1}{6.25} = -0.04 - 0.16 = -0.20$$

$$u_e = -5 \, \mathrm{cm}$$

So the objective's image lies $$5 \, \mathrm{cm}$$ in front of the eyepiece, i.e. at $$v_o = L - 5 = 15 - 5 = 10 \, \mathrm{cm}$$ from the objective.

For the objective: $$\dfrac{1}{u_o} = \dfrac{1}{v_o} - \dfrac{1}{f_o} = \dfrac{1}{10} - \dfrac{1}{2} = 0.1 - 0.5 = -0.4$$

$$u_o = -2.5 \, \mathrm{cm}$$

The object must be placed $$2.5 \, \mathrm{cm}$$ from the objective.

Magnifying power: $$m_o = \dfrac{v_o}{u_o} = \dfrac{10}{-2.5} = -4$$ and $$m_e = 1 + \dfrac{D}{f_e} = 1 + \dfrac{25}{6.25} = 5$$, so

$$|m| = |m_o| \times m_e = 4 \times 5 = 20$$

(b) Final image at infinity.

Now the intermediate image must lie exactly at the focus of the eyepiece, so $$u_e = -f_e = -6.25 \, \mathrm{cm}$$, giving $$v_o = L - 6.25 = 15 - 6.25 = 8.75 \, \mathrm{cm}$$.

$$\dfrac{1}{u_o} = \dfrac{1}{v_o} - \dfrac{1}{f_o} = \dfrac{1}{8.75} - \dfrac{1}{2} = 0.1143 - 0.5 = -0.3857$$

$$u_o = -2.59 \, \mathrm{cm}$$

The object must be placed about $$2.59 \, \mathrm{cm}$$ from the objective.

Magnifying power: $$m_o = \dfrac{v_o}{u_o} = \dfrac{8.75}{-2.59} = -3.38$$ and $$m_e = \dfrac{D}{f_e} = \dfrac{25}{6.25} = 4$$, so

$$|m| = 3.38 \times 4 \approx 13.5$$

Given $$f_o = 8.0 \, \mathrm{mm} = 0.8 \, \mathrm{cm}$$, $$f_e = 2.5 \, \mathrm{cm}$$, object distance $$u_o = -9.0 \, \mathrm{mm} = -0.9 \, \mathrm{cm}$$, near point $$D = 25 \, \mathrm{cm}$$.

Objective. Using $$\dfrac{1}{v_o} - \dfrac{1}{u_o} = \dfrac{1}{f_o}$$:

$$\dfrac{1}{v_o} = \dfrac{1}{f_o} + \dfrac{1}{u_o} = \dfrac{1}{0.8} + \dfrac{1}{-0.9} = 1.25 - 1.111 = 0.1389$$

$$v_o = \dfrac{1}{0.1389} = 7.2 \, \mathrm{cm}$$

Linear magnification of the objective: $$m_o = \dfrac{v_o}{u_o} = \dfrac{7.2}{-0.9} = -8$$.

Eyepiece. The person sees the final image at the near point, so $$v_e = -25 \, \mathrm{cm}$$.

$$\dfrac{1}{u_e} = \dfrac{1}{v_e} - \dfrac{1}{f_e} = \dfrac{1}{-25} - \dfrac{1}{2.5} = -0.04 - 0.4 = -0.44$$

$$u_e = -2.27 \, \mathrm{cm}$$

Separation between the lenses. The intermediate image lies $$v_o$$ beyond the objective and acts as the object $$|u_e|$$ in front of the eyepiece, so

$$L = v_o + |u_e| = 7.2 + 2.27 = 9.47 \, \mathrm{cm}$$

Magnifying power. $$m_e = 1 + \dfrac{D}{f_e} = 1 + \dfrac{25}{2.5} = 11$$, hence

$$m = |m_o| \times m_e = 8 \times 11 = 88$$

For a telescope in normal adjustment (final image at infinity), the magnifying power is the ratio of the focal lengths and the tube length is their sum.

Magnifying power.

$$m = \dfrac{f_o}{f_e} = \dfrac{144}{6.0} = 24$$

Separation between the lenses. In normal adjustment the image formed by the objective lies at the common focus of the two lenses, so the separation is

$$L = f_o + f_e = 144 + 6.0 = 150 \, \mathrm{cm}$$

When an object is viewed through a glass slab of thickness $$t$$ and refractive index $$n$$, refraction at the two parallel faces makes the object appear shifted towards the observer. For near-normal viewing the apparent shift (the distance by which the object appears raised) is

$$d = t\left( 1 - \dfrac{1}{n} \right)$$

With $$t = 15 \, \mathrm{cm}$$ and $$n = 1.5$$:

$$d = 15\left( 1 - \dfrac{1}{1.5} \right) = 15\left( 1 - 0.667 \right) = 15 \times \dfrac{1}{3} = 5 \, \mathrm{cm}$$

So the pin appears to be raised by $$5 \, \mathrm{cm}$$.

Does it depend on where the slab is? The expression $$d = t\left(1 - \dfrac{1}{n}\right)$$ involves only the thickness and refractive index of the slab — not the distance of the slab from the pin or from the eye. Hence the apparent shift is the same wherever the slab is held between the pin and the eye (for near-normal viewing). The given viewing distance of $$50 \, \mathrm{cm}$$ does not affect the result.

To project a real image of the bulb on the opposite wall, the lens must form a real image with the object (bulb) and image (wall) separated by the fixed distance $$D = 3 \, \mathrm{m}$$.

If the object is at a distance $$x$$ from the lens, the image is at $$D - x$$. For a real object and real image the lens formula gives

$$\dfrac{1}{D - x} + \dfrac{1}{x} = \dfrac{1}{f}$$

$$\dfrac{D}{x(D - x)} = \dfrac{1}{f} \implies f = \dfrac{x(D - x)}{D}$$

The focal length is largest when the product $$x(D - x)$$ is largest. This product is maximum at $$x = \dfrac{D}{2}$$, where $$x(D - x) = \dfrac{D^2}{4}$$. Hence

$$f_{\max} = \dfrac{D^2/4}{D} = \dfrac{D}{4} = \dfrac{3}{4} = 0.75 \, \mathrm{m}$$

Equivalently: a convex lens can form a real image only when the object–screen distance is at least $$4f$$. With $$D = 3 \, \mathrm{m}$$ this requires $$f \le 0.75 \, \mathrm{m}$$.

This is the displacement (conjugate-foci) method. For a fixed object–screen separation $$D$$, a convex lens forms a sharp real image at two positions; if those positions are a distance $$d$$ apart, the focal length is given by $$f = \dfrac{D^2 - d^2}{4D}$$.

Derivation. If the object distance is $$x$$, the image distance is $$D - x$$, and

$$\dfrac{1}{D - x} + \dfrac{1}{x} = \dfrac{1}{f}$$

leads to the quadratic $$x^2 - Dx + fD = 0$$. Its two roots $$x_1, x_2$$ are the two possible lens positions, with $$x_1 + x_2 = D$$ and $$x_1 x_2 = fD$$. Their separation is $$d = |x_1 - x_2|$$, so

$$d^2 = (x_1 + x_2)^2 - 4x_1 x_2 = D^2 - 4fD \implies f = \dfrac{D^2 - d^2}{4D}$$

Substituting $$D = 90 \, \mathrm{cm}$$ and $$d = 20 \, \mathrm{cm}$$:

$$f = \dfrac{(90)^2 - (20)^2}{4 \times 90} = \dfrac{8100 - 400}{360} = \dfrac{7700}{360} \approx 21.4 \, \mathrm{cm}$$

Let the ray strike the first face at angle of incidence $$i$$ and refract at angle $$r_1$$; inside the prism it then meets the second face at angle $$r_2$$. For a prism of refracting angle $$A$$,

$$r_1 + r_2 = A = 60^\circ$$

Condition at the second face. The ray just suffers total internal reflection there, which means $$r_2$$ equals the critical angle $$C$$:

$$\sin C = \dfrac{1}{n} = \dfrac{1}{1.524} = 0.6562 \implies C \approx 41.0^\circ$$

Therefore $$r_2 = 41.0^\circ$$, and

$$r_1 = A - r_2 = 60^\circ - 41.0^\circ = 19.0^\circ$$

Refraction at the first face. Snell's law gives $$\sin i = n \sin r_1$$:

$$\sin i = 1.524 \times \sin 19.0^\circ = 1.524 \times 0.3256 = 0.4962$$

$$i = \sin^{-1}(0.4962) \approx 29.75^\circ$$

So the ray must strike the first face at about $$29.75^\circ$$ for total internal reflection to just begin at the second face.

Data taken from Exercise 9.23

• Magnifying glass (simple microscope) of focal length  $$f = 10\,\text{cm}$$.
• The object is the small chequered card shown in Ex. 9.23; each little square on it has an actual area

$$A_o = 0.0625\,\text{mm}^2\;\;(\text{side }=0.25\,\text{mm}).$$

The present exercise demands that the area of the virtual image of one square be

$$A_i = 6.25\,\text{mm}^2 \;(\text{side }=2.50\,\text{mm}).$$

Hence the required linear magnification is

$$m = \frac{\sqrt{A_i}}{\sqrt{A_o}} = \frac{2.50}{0.25}=10.$$

1. Position of the object

For a thin lens (Cartesian sign–convention)

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f},\qquad m=\frac{v}{u}\;(m>1\text{ for an enlarged virtual image}).$$

Putting $$v = m u$$ and eliminating $$v$$:

$$\frac{1}{m u}-\frac{1}{u}=\frac{1}{f}\;\Rightarrow\;\frac{1-m}{m u}=\frac{1}{f}.$$

The object lies to the left of the lens, so $$u=-|u|$$ and $$(1-m)/(m u)=+(m-1)/(m|u|).$$
Thus

$$|u| = \frac{m-1}{m}\,f = \frac{10-1}{10}\,(10\,\text{cm}) = 9.0\,\text{cm}.$$

Therefore the object must be kept $$9\,\text{cm}$$ in front of the magnifying glass.

2. Will the eye resolve the squares?

The virtual image distance (from the lens) is

$$|v| = m|u| = 10\times9.0\,\text{cm}=90\,\text{cm}.$$

With the eye almost touching the lens, the image is also nearly $$90\,\text{cm}$$ from the eye. Its angular size is

$$\theta = \frac{\text{linear size}}{\text{distance}} = \frac{2.5\,\text{mm}}{900\,\text{mm}} \approx 2.8\times10^{-3}\,\text{rad} \approx 9.6\,\text{arc-min}.$$

The human eye can normally resolve detail down to about $$1\,\text{arc-min}$$, so $$\theta$$ is almost ten times larger. Hence the individual squares can indeed be seen distinctly with the eye kept very close to the magnifier.

We need a total magnifying power $$m = 30$$, with $$f_o = 1.25 \, \mathrm{cm}$$ and $$f_e = 5 \, \mathrm{cm}$$. Take the final image at the near point, $$D = 25 \, \mathrm{cm}$$ (the usual setting).

Split the magnification. The eyepiece contributes

$$m_e = 1 + \dfrac{D}{f_e} = 1 + \dfrac{25}{5} = 6$$

Since $$m = m_o \times m_e$$, the objective must give

$$m_o = \dfrac{m}{m_e} = \dfrac{30}{6} = 5$$

Object position for the objective. Here $$|m_o| = \dfrac{v_o}{|u_o|} = 5$$, so $$v_o = 5|u_o|$$. With $$u_o = -|u_o|$$ and $$v_o = +5|u_o|$$, the lens formula $$\dfrac{1}{v_o} - \dfrac{1}{u_o} = \dfrac{1}{f_o}$$ becomes

$$\dfrac{1}{5|u_o|} + \dfrac{1}{|u_o|} = \dfrac{1}{1.25} \implies \dfrac{6}{5|u_o|} = 0.8$$

$$|u_o| = \dfrac{6}{5 \times 0.8} = 1.5 \, \mathrm{cm}$$

So the object is placed $$1.5 \, \mathrm{cm}$$ from the objective, and the objective's image is at $$v_o = 5 \times 1.5 = 7.5 \, \mathrm{cm}$$.

Position of the eyepiece. The eyepiece must form the final image at $$v_e = -25 \, \mathrm{cm}$$:

$$\dfrac{1}{u_e} = \dfrac{1}{v_e} - \dfrac{1}{f_e} = \dfrac{1}{-25} - \dfrac{1}{5} = -\dfrac{6}{25}$$

$$u_e = -\dfrac{25}{6} \approx -4.17 \, \mathrm{cm}$$

Separation between the lenses.

$$L = v_o + |u_e| = 7.5 + 4.17 = 11.67 \, \mathrm{cm}$$

Hence: place the object $$1.5 \, \mathrm{cm}$$ in front of the objective, and set the two lenses about $$11.67 \, \mathrm{cm}$$ apart.

Given data

• Radius of curvature of the large (primary, concave) mirror
  $$R_1 = 220\,\text{mm}$$
• Radius of curvature of the small (secondary, convex) mirror
  $$R_2 = 140\,\text{mm}$$
• Distance between the poles of the two mirrors (secondary in front of primary)
  $$P_1P_2 = 20\,\text{mm}$$
• Object at infinity ⇒ parallel light falls on the primary mirror.

Step 1. Image produced by the primary mirror

For a spherical mirror

$$f = \dfrac{R}{2}$$

Using the Cartesian sign convention adopted in NCERT:

• Incident light travels from left to right, so distances measured to the right of a pole are positive, those to the left negative.
• The primary mirror is concave, its centre of curvature lies on the left of its pole ⇒ $$R_1 = -220\,\text{mm}$$

Hence

$$f_1 = \dfrac{R_1}{2} = -110\,\text{mm}$$

With the object at infinity, the image formed by the primary mirror is at its focus:

$$v_1 = f_1 = -110\,\text{mm}$$

Thus the primary image I₁ lies 110 mm to the left of the primary pole P₁.

Step 2. Object distance for the secondary mirror

The pole of the secondary mirror P₂ is 20 mm to the left of P₁. Hence the distance of I₁ from P₂ is

$$u_2 = P_2I_1 = 110\,\text{mm} - 20\,\text{mm} = 90\,\text{mm}$$

The rays that have already been reflected by the primary now travel from right to left towards the secondary; therefore, according to the sign convention, distances measured to the left of P₂ are positive. Hence

$$u_2 = +90\,\text{mm}$$

Step 3. Image produced by the secondary mirror

• The secondary mirror is convex; its centre of curvature is to the right of P₂, so $$R_2 = +140\,\text{mm} \;\Rightarrow\; f_2 = \dfrac{R_2}{2} = +70\,\text{mm}$$. For the convex mirror, however, the focal length is negative in the Cartesian convention because it lies on the opposite side to the incident light. Thus

$$f_2 = -70\,\text{mm}$$

Mirror formula:

$$\frac{1}{f_2}=\frac{1}{v_2}+\frac{1}{u_2}$$

$$\frac{1}{(-70)}=\frac{1}{v_2}+\frac{1}{(+90)}$$

$$\frac{1}{v_2}=\frac{-1}{70}-\frac{1}{90}=\frac{-9-7}{630}=\frac{-16}{630}$$

$$v_2 = -\frac{630}{16}\,\text{mm}\approx -39.4\,\text{mm}$$

The negative sign shows that the image formed by the secondary mirror lies to the right of P₂, i.e. on the same side as the hole in the primary mirror.

Step 4. Position of the final image with respect to the primary mirror

Distance of the final image I₂ from the pole of the primary mirror:

$$P_1I_2 = P_1P_2 + P_2I_2 = 20\,\text{mm} + 39.4\,\text{mm} \approx 59.4\,\text{mm}$$

This is to the right of the primary mirror.

Result

The Cassegrain combination forms the final image approximately 59 mm behind the large mirror (on the side opposite to the incoming light).

Hence, an object at infinity is brought to focus about 6 cm behind the primary mirror.

The light is initially incident normally on the plane mirror, so the incident ray and the reflected ray coincide and move along the same line.

When a current passes through the galvanometer coil, the mirror (and hence its normal) rotates through an angle $$\theta = 3.5^{\circ}$$. According to the law of reflection, the angle between the incident ray and the new normal increases by $$\theta$$, and the angle between the reflected ray and the old direction of the incident ray therefore changes by

$$\Delta = 2\theta$$

This means the reflected beam is turned through $$2\theta$$ with respect to its original direction.

Distance of the screen from the mirror: $$L = 1.5\,\text{m}$$.

Linear displacement of the spot on the screen:

$$s = L\,\tan(2\theta)$$

Convert the deflection angle to radians:

$$\theta = 3.5^{\circ} = 3.5\times\frac{\pi}{180}\,\text{rad} = 0.0611\,\text{rad}$$

Hence

$$2\theta = 0.1222\,\text{rad} \;(= 7^{\circ})$$

Now,

$$\tan(2\theta) = \tan 7^{\circ} \approx 0.123$$

Therefore

$$s = 1.5\,\text{m}\;\times 0.123 \approx 0.184\,\text{m}$$

$$s \approx 1.8\times 10^{-1}\,\text{m} = 18\,\text{cm}$$

The reflected spot shifts by about 18 cm on the screen.

Step 1 : Focal length of the equiconvex lens in air
With the liquid removed the lens is in contact with the plane mirror. A ray emerging from the lens immediately strikes the mirror and retraces its path after reflection, i.e. the lens is traversed twice. Hence the optical power doubles and the combination behaves like a concave mirror of focal length $$F_0 = \frac{f}{2}$$ where $$f$$ is the focal length of the lens (in the surrounding medium – here air).
For an object placed at the centre of curvature of a concave mirror the image is real, inverted and coincident with the object; therefore $$2F_0 = f = 30\;\text{cm}.$$ Thus

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

Step 2 : Radius of curvature of the lens surfaces
For a thin equiconvex glass lens (refractive index $$n_g = 1.50$$) in air (R₁ = +R, R₂ = -R) the lens-maker’s formula gives $$\frac{1}{f}=\left(n_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) =(1.50-1)\left(\frac{1}{R}-\frac{-1}{R}\right)=0.5\;\frac{2}{R}=\frac{1}{R}.$$ Consequently

$$\boxed{R = 30\;\text{cm}}$$

Step 3 : Focal length of the same lens when its right surface faces a medium of refractive index $$\mu$$ (the liquid)
Let the incident medium be air (index $$n_1 = 1$$), the lens glass index $$n_g = 1.50$$ and the emergent medium the liquid (index $$\mu$$). For a thin lens separating two different media the power (in dioptres) is $$P = \left(\frac{n_g-n_1}{R_1}+\frac{\mu-n_g}{R_2}\right).$$ Putting R₁=+R, R₂=−R we obtain $$P = \frac{1.50-1}{R}+\frac{\mu-1.50}{-R}=\frac{0.50-\mu+1.50}{R}=\frac{2-\mu}{R}.$$ The focal length (measured in air on the object side) is therefore $$\boxed{f_1 = \frac{1}{P}=\frac{R}{2-\mu}}\qquad(1)$$

Step 4 : Condition for coincidence of object and (inverted) image
With the liquid in place the lens is again traversed twice; its power doubles and the combination acts as a concave mirror of focal length $$F = \frac{f_1}{2}.$$ For the object to coincide with its inverted image it must be placed at the centre of curvature of this equivalent mirror: $$2F = f_1.$$ The experiment gives the required object distance (from the lens) $$f_1 = 45\;\text{cm}.\qquad(2)$$

Step 5 : Determination of the refractive index of the liquid
Substituting $$R = 30\;\text{cm}$$ and $$f_1 = 45\;\text{cm}$$ in (1): $$45 = \frac{30}{2-\mu} \;\;\Rightarrow\;\;2-\mu = \frac{30}{45}=\frac{2}{3} \;\;\Rightarrow\;\;\boxed{\mu = \frac{4}{3}\;\;\approx\;\;1.33}.$$

The liquid therefore has a refractive index very close to that of water.