Chapter 10: The Human Eye and the Colourful World

Step 1 : Why the eye must adjust itself
For an object to be seen clearly, its image must be formed exactly on the retina. The distance between the eye-lens and the retina is fixed (about $$2.5\,\text{cm}$$). However, the objects we look at lie at very different distances — from the near point (about $$25\,\text{cm}$$) right up to the far point (infinity). To keep the image always on the retina, the focal length of the eye-lens itself must change.

Step 2 : How the focal length is changed
The eye-lens is made of soft, elastic, jelly-like material. It is held in place by the ciliary muscles. When these muscles relax, the lens becomes thin and its focal length increases — this allows distant objects to be focused. When the muscles contract, the lens becomes thicker and more curved, its focal length decreases, and nearby objects can be focused.

Step 3 : Definition
The power of accommodation of the eye is the ability of the eye-lens to adjust its focal length (by the action of the ciliary muscles) so that the images of objects lying at different distances — from the near point ($$\approx 25\,\text{cm}$$) to the far point (infinity) — are always formed sharply on the retina.

However, the focal length of the eye-lens cannot be reduced below a certain minimum value. This is why objects placed closer than about $$25\,\text{cm}$$ from the eye cannot be seen clearly.

Given data

• The far point of the myopic eye = $$1.2\,\text{m}$$ from the eye.
• To see a very distant object (effectively at infinity) clearly, the corrective lens must form its image at the far point of the eye.

Lens formula

For a thin lens we use $$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$$, taking the usual Cartesian sign convention (all distances are measured from the lens; left of the lens is negative, right is positive).

Object distance : a distant object is at infinity, so $$u=-\infty$$.

Image distance : the image has to be produced at the eye’s far point, which is on the same side of the lens as the object (to the left). Hence $$v=-1.2\,\text{m}$$.

Substituting in the lens formula,

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

$$\frac{1}{-1.2\,\text{m}}-\frac{1}{-\infty}=\frac{1}{f}$$

Since $$\frac{1}{-\infty}=0$$, we get

$$\frac{1}{f}=\frac{1}{-1.2\,\text{m}}=-0.833\,\text{m}^{-1}$$

Therefore, $$f=-1.2\,\text{m}$$.

Power of the lens

Power $$P$$ (in dioptres) is $$P=\dfrac{1}{f\,(\text{in m})}$$, so

$$P=\frac{1}{-1.2}=-0.833\,\text{D}\;(\text{≈}-0.83\,\text{D}).$$

Nature of the lens

A negative focal length and negative power correspond to a diverging (concave) lens.

Conclusion

To correct the defect, the person should use a concave lens of focal length $$-1.2\,\text{m}$$ (power about $$-0.83\,\text{D}$$).

For a lens system such as the eye, an object is seen clearly only when its image is formed exactly on the retina. The position(s) of the object for which this happens without the eye having to apply any extra strain are called the least distance of distinct vision (near point) and the far point.

Near point (least distance of distinct vision)
The nearest position of an object from the eye that can be brought to sharp focus on the retina by maximum accommodation (ciliary muscles fully contracted) is called the near point. For a healthy adult with normal ("emmetropic") vision, extensive experiments give

$$D = 25\,\text{cm}$$

Thus the normal human eye cannot see an object kept closer than 25 cm distinctly.

Far point
The farthest position of an object from the eye that can be seen clearly when the eye is completely relaxed (no accommodation) is called the far point. For a normal eye this distance is effectively infinite; any object at a distance of 6 m or more sends light rays to the eye that are practically parallel, and the eye focuses them on the retina without effort. Hence we write

$$S = \infty$$

Conclusion
For a human eye with normal vision:

• Near point, $$D = 25\,\text{cm}$$
• Far point, $$S = \infty$$ (practically anything beyond about 6 m)

The information given is:

• The child reads from the last row (so the object – the blackboard writing – is at a large distance, practically at infinity).
• He/She cannot see this distant object clearly.

This is the typical symptom of myopia (near-sightedness): a person can see nearby objects distinctly but distant objects appear blurred.

Reason for the defect

• The eye-lens system has too much converging power; or
• The eyeball has become elongated in the forward–backward direction.

Because of either (or both) of these, parallel rays from a distant object are brought to focus at a point $$I$$ in front of the retina, so the retina receives a blurred image.

Ray-diagram to draw (describe in your notebook):

• Draw an eye with its lens.
• Mark the retina further back.
• Show parallel rays from the blackboard being focused before the retina.
• After placing a concave lens in front of the eye, show the rays first diverging, then being focused exactly on the retina.

Optical correction

Place a concave (diverging) lens of suitable focal length $$f$$ in front of the eye.

• Let the far point of the myopic eye be at distance $$D$$ from the eye (for a normal eye $$D = \infty$$).
• The corrective lens must form the image of an object at infinity at this far point, so for the lens we have the lens formula

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

• For an object at infinity $$u = -\infty \; (\text{taking the usual sign convention})$$, so $$1/u \approx 0$$.
• The image must be at the far point, so $$v = -D$$ (negative because it is on the same side as the object for a concave lens).

Hence $$ \frac{1}{f} = 0 - \frac{1}{(-D)} = -\frac{1}{D} $$

$$ \Rightarrow \; f = -D $$   (negative focal length ⇒ concave lens)

The power required is therefore  $$ P = \frac{1}{f\,(\text{in m})} = -\frac{1}{D\,(\text{in m})} \;\text{dioptre (D)} $$

Thus a concave lens of power $$P$$ brings the focus back onto the retina, and the child can read the blackboard clearly.

Conclusion

The child is suffering from myopia. It is corrected by wearing spectacles containing a concave (diverging) lens of suitable focal length/power so that the image of distant objects is formed on the retina.

The eye behaves like a camera whose lens–screen distance (≈ distance between the eye–lens and retina) is fixed. When an object is moved closer to the eye, the image has to remain sharply on the retina. To achieve this, the eye must decrease the focal length of its lens; when the object is taken far away, the focal length must again increase. This continuous change in focal length is obtained as follows:

• The ring-shaped ciliary muscles contract or relax.
• This changes the curvature (and therefore the focal length $$f$$) of the elastic eye-lens:

• For a nearby object  ⇒  muscles contract  ⇒  lens becomes thicker  ⇒  smaller $$f$$.
• For a distant object  ⇒  muscles relax  ⇒  lens becomes thinner  ⇒  larger $$f$$.

The ability of the eye to vary its focal length so that images of objects at different distances are always formed on the retina is called accommodation.

Why the other options are incorrect:

1. Presbyopia is an age-related defect in which accommodation itself becomes weak.
2. Near-sightedness (myopia) is a defect in which distant objects cannot be clearly seen.
3. Far-sightedness (hypermetropia) is a defect in which nearby objects are not seen clearly.

Hence the correct choice is option (b).

Light from an object successively crosses the following parts of the eye:

1. Cornea – the transparent, curved membrane that does the maximum refraction.
2. Aqueous humour – a clear fluid behind the cornea.
3. Pupil – the adjustable opening through which light enters; its size is controlled by the coloured iris.
4. Eye-lens – provides the remaining convergence needed to focus the rays.

After refraction by the lens, the rays converge on the light-sensitive screen that lines the back wall of the eyeball, called the retina. A real, inverted image of the object is therefore formed on the retina, and the optic nerve then conveys the visual information to the brain.

Thus, the image of an object in the human eye is formed at the retina.

Correct option: (d) retina.

The eye cannot focus on objects that are brought very close to it.

The smallest distance at which a normal (emmetropic) eye can see an object clearly and comfortably is called the least distance of distinct vision or the near point.

For a young adult with normal vision, experimental observations give

$$D = 25\;\text{cm} = 0.25\;\text{m}.$$

Thus the value closest to the standard near-point distance is option (c).

For an object at different distances, the eye must be able to change the focal length of its eye-lens so that the image is always formed on the retina. This ability is called accommodation.

The parts listed in the options and their roles are:

• Pupil: An opening whose size is controlled by the iris to regulate the intensity of light entering the eye; it does not alter focal length.
• Retina: The light-sensitive screen where the image is formed; it plays no part in changing focal length.
• Ciliary muscles: A ring of muscles attached to the eye-lens. By contracting or relaxing, they change the curvature (and hence focal length) of the lens, enabling accommodation.
• Iris: The coloured diaphragm around the pupil that controls the pupil’s size; like the pupil, it does not affect focal length.

Therefore, the change in focal length of the eye-lens is caused by the action of the ciliary muscles.

Given data

• Power needed to correct distant vision: $$P_d = -5.5\,\text{D}$$
• Power needed to correct near vision: $$P_n = +1.5\,\text{D}$$

Basic relation between power and focal length

For any thin lens, the optical power $$P$$ (in dioptres) and the focal length $$f$$ (in metres) are related by

$$P = \frac{1}{f}$$  or  equivalently  $$f = \frac{1}{P}$$

Note : A negative power means a negative focal length (concave or diverging lens); a positive power means a positive focal length (convex or converging lens).

(i) Focal length for distant‐vision correction

Substitute $$P_d = -5.5\,\text{D}$$ into $$f = 1/P$$:

$$f_d = \frac{1}{-5.5}\,\text{m}$$

$$f_d = -0.1818\,\text{m}$$

Convert to centimetres (1 m = 100 cm):

$$f_d = -0.1818\times100\,\text{cm} = -18.18\,\text{cm}$$

So, a concave lens of focal length about $$-18\,\text{cm}$$ is required.

(ii) Focal length for near‐vision correction

Substitute $$P_n = +1.5\,\text{D}$$ into $$f = 1/P$$:

$$f_n = \frac{1}{+1.5}\,\text{m}$$

$$f_n = 0.6667\,\text{m}$$

Convert to centimetres:

$$f_n = 0.6667\times100\,\text{cm} = 66.67\,\text{cm}$$

Thus, a convex lens of focal length about $$+67\,\text{cm}$$ is required.

Step 1 | Understand the situation
The far point of the myopic eye is only $$80\,\mathrm{cm}$$ from the eye. For clear distant vision the correcting lens must form the image of an object at infinity at this far point.

Step 2 | Sign convention and known quantities
Taking the usual Cartesian sign convention:

• Object distance for an object at infinity: $$u = -\infty$$ (to the left of the lens).
• Required image distance (at the far point, in front of the lens): $$v = -80\,\text{cm} = -0.80\,\text{m}$$.

Step 3 | Apply the lens formula
The thin-lens formula is $$\dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f}$$.

Substituting the values:

$$\frac{1}{f} = \frac{1}{(-0.80)} - \frac{1}{(-\infty)} = -\frac{1}{0.80} - 0 = -1.25\,\text{m}^{-1}$$

Hence $$f = -0.80\,\text{m}$$.

Step 4 | Determine the power
Power of a lens: $$P = \dfrac{1}{f\,(\text{in m})}$$

$$P = \frac{1}{-0.80} = -1.25\,\text{dioptres (D)}$$

Step 5 | Interpret the sign
The negative focal length and negative power indicate a diverging (concave) lens.

Result: A concave lens of power $$-1.25\,\mathrm{D}$$ (focal length $$-0.80\,\mathrm{m}$$) is required to correct the myopia.

Step 1 – Understand the defect
In hypermetropia (farsightedness) the eyeball is too short or the eye lens is too weak. Light rays coming from a nearby object focus behind the retina, so the person cannot see nearby objects clearly.

Step 2 – Diagram to show correction
Draw the following labelled diagram on paper:

• A horizontal principal axis.
• On the right, sketch an eyeball with its eye-lens and retina.
• Mark the focal point of the uncorrected eye behind the retina to indicate hypermetropia.
• Place a thin convex spectacle lens in front of the eye on the principal axis.
• On the left of the spectacle lens draw an object (say a book) at 25 cm from the lens.
• Show two incident rays from the top of the object striking the convex lens, refracting towards the principal axis and then entering the eye-lens such that their backward extensions meet at a point 1 m from the lens on the object side (virtual image). This virtual image lies at the eye’s near point, so the defective eye can now focus it exactly on the retina.

Step 3 – Data given
Near point of the defective (hypermetropic) eye: $$D = 1\,\text{m}$$.
Near point of a normal eye: $$d_n = 25\,\text{cm} = 0.25\,\text{m}$$.

Step 4 – Decide what the correcting lens must do
When the person holds a book at 25 cm, the lens must form a virtual image of the book at the eye’s farthest near-point, i.e. 1 m in front of the eye, so that the eye can focus it comfortably.

Step 5 – Sign convention
For a thin lens (cartesian convention):

• Object is placed to the left of the lens → $$u$$ is negative.
• Virtual image also appears on the left → $$v$$ is negative.

Step 6 – Substitute in the lens formula

Object distance: $$u = -0.25\,\text{m}$$
Image distance: $$v = -1.0\,\text{m}$$

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

Calculate: $$\frac{1}{f} = \frac{1}{-1.0} - \frac{1}{-0.25} = -1 + 4 = 3\;\text{m}^{-1}$$

Therefore, $$f = \frac{1}{3}\,\text{m} = 0.333\,\text{m}$$ (positive → convex lens).

Step 7 – Power of the lens
Power $$P$$ of a lens in dioptres: $$P = \frac{1}{f\,(\text{in m})}$$

Thus, $$P = \frac{1}{0.333} \approx 3\;\text{D}$$.

Result
A convex lens of power $$+3\;\text{D}$$ (focal length about 33 cm) is required to correct the given hypermetropic eye.

Near point and accommodation

The human eye can change the focal length of its eye lens by tightening or relaxing the ciliary muscles. This adjustment, called accommodation, enables the eye to form a sharp image of an object on the retina whose distance from the optical centre of the eye (O) is fixed at about $$v \approx 2.5\,\text{cm}$$.

The closest point that a normal (healthy) eye can accommodate without strain is known as the near point. For a normal adult this distance is $$25\,\text{cm}$$ in front of the eye.

Lens-formula calculation

The formation of the retinal image obeys the lens formula

$$\frac1f = \frac1v - \frac1u$$

where

• $$u$$ = object distance (taken negative because the object is in front of the lens)
• $$v$$ = image distance $$\approx +2.5\,\text{cm}$$ (positive, measured inside the eye)
• $$f$$ = focal length of the eye lens (to be adjusted by the ciliary muscles)

For an object placed at the near point, $$u = -25\,\text{cm}$$:

$$\frac1f = \frac1{2.5} - \frac1{(-25)} = 0.40 + 0.04 = 0.44\,\text{cm}^{-1}$$

$$\Rightarrow\; f_{\min} \approx 2.27\,\text{cm}$$

This is the shortest focal length a normal eye can achieve comfortably. If the object is brought still closer, say to $$u = -15\,\text{cm}$$, the required focal length becomes

$$\frac1f = 0.40 + 0.0667 = 0.4667\,\text{cm}^{-1} \;\;\Rightarrow\; f \approx 2.14\,\text{cm}$$

The eye lens would have to become still more curved (smaller $$f$$). Beyond a reduction to about $$2.27\,\text{cm}$$ the ciliary muscles cannot increase the curvature any further without extreme strain. Consequently the image can no longer be brought precisely on the retina and appears blurred.

Conclusion

Because the ciliary muscles of a normal eye cannot decrease the focal length of the eye lens below roughly $$2.27\,\text{cm}$$, objects kept closer than $$25\,\text{cm}$$ cannot be focused sharply on the retina. Hence such objects are seen blurred by a normal eye.

Concept used : A healthy human eye forms the image of every object on the retina. The position of the retina from the eye’s optical centre is fixed (about 2.5 cm). Hence, while viewing objects at different distances, the eye cannot shift the screen; instead, it changes the focal length of its lens (accommodation).

Lens-maker’s equation for the eye

For a thin lens in air we may treat the eye as obeying the lens formula

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

• $$u$$ = object distance (measured from the optical centre of the eye; it is negative for a real object in the Cartesian sign convention).
• $$v$$ = image distance = distance of the retina from the eye’s lens (≈ 2.5 cm and fixed).
• $$f$$ = focal length of the eye lens (variable; changed by ciliary muscles).

What happens when the object is moved farther away?

1. Increasing the object distance means $$|u|$$ increases, so $$\frac{1}{u}$$ decreases in magnitude.
2. Because the retina cannot move, $$v$$ must stay the same; therefore $$\frac{1}{v}$$ is constant.
3. To keep $$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$$ valid, a decrease in $$\frac{1}{u}$$ must be balanced by a decrease in $$\frac{1}{f}$$.
4. A smaller $$\frac{1}{f}$$ means a larger focal length $$f$$. Thus the eye lens becomes thinner (its curvature reduces).

Conclusion

The image distance in the eye does not change; it remains equal to the fixed distance between the eye lens and the retina. What changes instead is the focal length of the lens to maintain a sharp image.

Step 1 : Treat the star as a point-like source at a very large distance
The angular size of a star is so small that, for an observer on the Earth, the light can be assumed to come from an ideal point object. Hence even a slight change in the path or intensity of its rays is noticeable.

Step 2 : Recall atmospheric refraction
The Earth’s atmosphere is not uniform. From top to bottom the air becomes denser, so its refractive index increases from $$n_0$$ (nearly 1) in the upper, rare layers to $$n$$ (>$$n_0$$) near the surface.
Whenever a ray of light travels from a medium of refractive index $$n_1$$ to another of refractive index $$n_2$$, Snell’s law applies: $$n_1\sin i = n_2\sin r$$.
Because $$n_2 > n_1$$ for a downward ray, $$\sin r < \sin i$$, i.e. the ray bends towards the normal. Thus starlight is gradually bent towards the normal as it comes down through successive layers.

Step 3 : Recognise that the atmosphere is not steady
Temperature, pressure and wind keep changing every moment. Hence each layer’s refractive index fluctuates by a small amount $$\Delta n$$. Therefore the actual path of the ray and its final angle of incidence at the eye keep varying continuously.

Step 4 : Fluctuating path leads to two simultaneous effects

1. Apparent position shift: Every small change $$\Delta n$$ changes the emergent direction by $$\Delta \theta$$ (obtained from differentiating Snell’s law). The star therefore seems to dance about.
2. Intensity fluctuation: When the refracted beam slightly misses the eye, the intensity falls; when it enters the eye completely, the intensity rises. The received intensity $$I(t)$$ therefore varies with time.

These rapid and random variations of $$I(t)$$ give the familiar bright-dim-bright pattern called twinkling.

Step 5 : Why planets do not twinkle
Planets are much closer, so they have a measurable angular size. Light from a planet reaches the eye in many closely spaced, parallel rays. Each ray suffers independent fluctuations, but their intensities add up. The statistical averaging smooths out the variations, keeping the total intensity nearly constant; hence planets usually shine with a steady light.

Conclusion
Because a star is effectively a point source and because the Earth’s turbulent atmosphere keeps changing its refractive index, both the apparent direction and the intensity of starlight reaching the eye fluctuate rapidly. The net result is that stars appear to twinkle.

Known facts from the chapter

• Stars and planets are both luminous celestial objects, but they differ greatly in their apparent (angular) size when seen from the Earth.
• Light coming from any celestial object has to pass through the Earth’s atmosphere, whose optical density is not uniform. The atmosphere is made of layers of air that are in continuous random motion; each layer has a slightly different temperature and therefore a slightly different refractive index.
• This non-uniform, turbulent atmosphere produces atmospheric refraction. When the refractive index of the medium through which light travels keeps fluctuating, the direction and intensity of the light reaching our eye also fluctuate.

Step 1 — Angular size comparison

• The average angular (apparent) diameter of a bright star, such as Sirius, is far below $$1''$$ (one arc-second). For all practical purposes, a star behaves like a point source.
• Typical planets (e.g. Venus, Jupiter, Mars) have much larger apparent diameters, from about $$5''$$ up to more than $$50''$$ depending on their distance from Earth. In other words, a planet is an extended source consisting of thousands of point-like spots packed together in a tiny disk.

Step 2 — Effect of atmospheric refraction on a point source (star)

• Because a star is nearly a point, a small change in the refractive index along its line of sight is enough to deflect the entire beam entering the eye by a tiny angle $$\delta\theta$$.
• When $$\delta\theta$$ is comparable to or larger than the star’s own apparent diameter (< $$1''$$), the star’s image shifts randomly about its mean position. At the same time, some rays are bent away from the pupil while others are bent into it, so the brightness (intensity) also varies. This rapid fluctuation of position and brightness is perceived as twinkling.

Step 3 — Effect of atmospheric refraction on an extended source (planet)

• A planet presents a small disk made of many individual point-like spots. Light from one spot might be deviated slightly upward, another spot’s light might be deviated slightly downward, and so on.
• Because the directions of deviation for the multitude of spots differ independently, the net effect averages out.
• Mathematically, if the disk of the planet subtends an angle $$\theta_p$$ at the eye and the typical random deviation of any ray is $$\delta\theta$$, then the fractional deviation in total light entering the pupil is roughly $$\dfrac{\delta\theta}{\theta_p}$$.
• For stars, $$\theta_p \approx 0\,\text{(point)} \Longrightarrow \dfrac{\delta\theta}{\theta_p}$$ becomes large, hence strong fluctuation.
  For planets, $$\theta_p$$ is 10 to 100 times larger than $$\delta\theta$$, so $$\dfrac{\delta\theta}{\theta_p} \ll 1$$; brightness remains practically constant.

Step 4 — Conclusion in textbook language

Planets do not twinkle because:

1. They are much closer to Earth and therefore have a comparatively larger apparent size (extended sources).
2. The atmospheric refraction effects on the different points of the planetary disk cancel one another out; the eye receives a steady average intensity.
3. Hence the position and brightness of a planet remain almost unchanged, so no twinkling is observed.

Step 1 : Recall the reason why the sky looks blue from the ground
When sunlight enters the Earth’s atmosphere it meets molecules of air, tiny dust and water droplets. Because these particles are much smaller than the wavelength of visible light, the relevant mechanism is Rayleigh scattering.

For Rayleigh scattering the scattered intensity is inversely proportional to the fourth power of wavelength:

$$I \propto \dfrac{1}{\lambda^{4}}$$

Short-wavelength blue/violet light (smaller $$\lambda$$) is therefore scattered far more strongly than long-wavelength red light. The scattered blue light reaches our eyes from every direction, so we perceive the daytime sky as blue.

Step 2 : Conditions faced by an astronaut
An astronaut orbiting the Earth is essentially in a vacuum. Outside the thin layer of atmosphere no significant medium (air molecules, dust, water droplets, etc.) is present.

Step 3 : Consequence of the absence of a scattering medium
Because $$I \propto 1/\lambda^{4}$$ holds only when there is a medium, in near-vacuum the value of $$I$$ for every wavelength becomes practically zero. With nothing to scatter sunlight, no diffuse light enters the astronaut’s eyes from surrounding directions.

Step 4 : Resulting appearance of the sky
The astronaut sees direct sunlight only along the line of sight to the Sun; in every other direction there is no light, so the sky looks completely dark — nearly black — instead of blue.

Hence, the sky appears dark (black) to an astronaut because there is virtually no atmosphere in space to scatter sunlight.