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The eye uses a lens system to form an image on a light-sensitive screen called the retina. The eyeball is nearly spherical with a diameter of about 2.3 cm.
The eye lens forms a real and inverted image of the object on the retina. The brain processes the signals and we perceive the object upright.
The eye lens is made of a jelly-like fibrous material. Its curvature can be changed by the ciliary muscles, which changes its focal length. The ability of the eye lens to adjust its focal length is called accommodation.
In old age the crystalline lens can become milky and cloudy. This is cataract, and it causes partial or complete loss of vision. Vision can be restored by cataract surgery.
When the eye loses its power of accommodation, vision becomes blurred. The three common refractive defects are myopia, hypermetropia and presbyopia. They are corrected with suitable spherical lenses.
With age the power of accommodation decreases, the ciliary muscles weaken and the lens becomes less flexible. The near point recedes, and reading needs corrective glasses.
A person with both myopia and hypermetropia may need bi-focal lenses: the upper part is concave for distant vision and the lower part is convex for near vision. Defects can also be corrected with contact lenses or surgery.
The power of a lens is the reciprocal of its focal length in metres. Its unit is the dioptre (D). A convex lens has positive power and a concave lens has negative power.
P = 1 / fA pair of donated eyes can give vision to up to four corneal-blind people. Many cases of blindness can be cured by corneal transplantation.
A triangular glass prism has two triangular bases and three rectangular lateral faces. The angle between its two refracting (lateral) faces is the angle of the prism.
On entering the first face, from air into glass, the ray bends towards the normal. On leaving the second face, from glass into air, it bends away from the normal. Because the two faces are inclined to each other, the emergent ray is not parallel to the incident ray. In a rectangular glass slab the emergent ray is parallel to the incident ray and only shifted sideways.
The angle between the direction of the incident ray and the direction of the emergent ray is the angle of deviation (D).
A prism splits a narrow beam of white light into a band of colours, called the spectrum. The splitting of light into its component colours is called dispersion.
Isaac Newton first obtained the spectrum of sunlight with a prism. A second similar prism could not split the colours any further. When he placed a second identical prism in an inverted position, the colours recombined into white light. This showed that sunlight is made of seven colours.
A rainbow is a natural spectrum appearing in the sky after a rain shower. Tiny water droplets act like small prisms: they refract and disperse sunlight, reflect it internally, and refract it again as it leaves the drop. A rainbow is always formed in the direction opposite to the Sun, and can also be seen near a waterfall or fountain with the Sun behind you.
Objects seen through hot air rising above a fire or radiator seem to waver. Hot air is less dense and has a slightly lower refractive index than cooler air above it, and since this air keeps moving, the apparent position of the object keeps changing.
Starlight passes through layers of air of gradually changing refractive index and bends towards the normal each time. The star therefore appears slightly higher than its actual position. Because the air is not steady, the apparent position keeps changing and the amount of light entering the eye keeps varying. The star seems brighter at one instant and fainter at the next: this is twinkling. Stars are so far away that they act as point sources.
Planets are much closer and behave like extended sources, a collection of many point sources. The variations in light from these points average out, so the total amount of light entering the eye stays nearly steady.
Because of atmospheric refraction the Sun is seen about 2 minutes before the actual sunrise and about 2 minutes after the actual sunset. The same effect makes the Sun's disc look flattened near the horizon.
The path of a beam of light becomes visible when it passes through a colloidal solution or a mixture with fine particles, such as smoke, mist, dust or water droplets. This scattering of light by colloidal particles is the Tyndall effect. It is seen when a fine beam of sunlight enters a smoke-filled room or passes through a dense forest canopy.
The colour of scattered light depends on the size of the particles. Very fine particles scatter mainly blue light. Larger particles scatter longer wavelengths too, and if they are large enough the scattered light appears white.
Air molecules and other fine particles in the atmosphere are smaller than the wavelength of visible light. They scatter short wavelengths (blue) much more than long wavelengths (red); red light has a wavelength about 1.8 times that of blue. The scattered blue light reaches our eyes, so the clear sky looks blue.
Without an atmosphere there would be no scattering and the sky would look dark. At very high altitudes, where scattering is not prominent, the sky appears dark to passengers.
Red light is scattered the least by fog or smoke, so it can be seen from far away in the same colour. This is why danger signals are red.
Near the horizon, sunlight travels a longer path through the atmosphere. Most of the blue and shorter wavelengths are scattered away, and the light that reaches us is mostly red, so the Sun looks reddish.
| Defect | Cannot see clearly | Image forms | Lens used |
|---|---|---|---|
| Myopia | Distant objects | In front of the retina | Concave |
| Hypermetropia | Nearby objects | Behind the retina | Convex |
| Presbyopia | Nearby objects (old age) | Behind the retina | Convex (or bi-focal) |
Try each problem first, then open the solution to check your working.
A convex lens has a focal length of 50 cm. Find its power.
Given: f = +50 cm = +0.5 m
Use: P = 1 / f
P = 1 / 0.5 = +2 D
Answer: +2 D
A concave lens has a focal length of 25 cm. Find its power.
Given: f = −25 cm = −0.25 m
Use: P = 1 / f
P = 1 / (−0.25) = −4 D
Answer: −4 D
A lens has a power of +2.5 D. Find its focal length and say what type of lens it is.
Given: P = +2.5 D
Use: f = 1 / P
f = 1 / 2.5 = 0.4 m = 40 cm
The power is positive, so it is a convex lens.
Answer: f = +40 cm; convex
A lens has a power of −0.8 D. Find its focal length.
Given: P = −0.8 D
Use: f = 1 / P
f = 1 / (−0.8) = −1.25 m = −125 cm
Answer: −125 cm (concave)
A +3 D lens and a −1 D lens are placed in contact. Find the power and focal length of the combination.
Given: P₁ = +3 D; P₂ = −1 D
Use: P = P₁ + P₂
P = 3 + (−1) = +2 D
f = 1 / 2 = 0.5 m = 50 cm
Answer: +2 D; f = 50 cm
A myopic person cannot see objects beyond 1.2 m distinctly. What lens is needed, and what is its power?
Given: Far point = 1.2 m
Use: The lens must form a virtual image of a distant object at the far point, so f = −(far point)
f = −1.2 m
P = 1 / f = 1 / (−1.2) = −0.83 D
A negative power means a concave lens.
Answer: Concave lens of power −0.83 D
The far point of a myopic person is 80 cm in front of the eye. Find the nature and power of the corrective lens.
Given: Far point = 80 cm = 0.8 m
Use: P = 1 / f, with f = −0.8 m
P = 1 / (−0.8) = −1.25 D
Answer: Concave lens, −1.25 D
A student's far point is 2 m. What is the power of the lens needed?
Given: Far point = 2 m
Use: P = −1 / (far point)
P = −1 / 2 = −0.5 D
Answer: −0.5 D (concave)
The spectacles of a myopic student have a power of −2.5 D. What is the student's far point?
Given: P = −2.5 D
Use: f = 1 / P and far point = |f|
f = 1 / (−2.5) = −0.4 m
The far point is 0.4 m.
Answer: 40 cm
A person needs −5.5 D for distant vision and +1.5 D for near vision. Find the focal length of each lens.
Given: P₁ = −5.5 D; P₂ = +1.5 D
Use: f = 1 / P
f₁ = 1 / (−5.5) = −0.182 m = −18.2 cm (concave, for distant vision)
f₂ = 1 / 1.5 = 0.667 m = +66.7 cm (convex, for near vision)
Answer: −18.2 cm and +66.7 cm
The near point of a hypermetropic eye is 1 m. What is the power of the lens needed? Take the normal near point as 25 cm.
Given: Object distance u = −25 cm; image distance v = −100 cm (virtual image on the same side)
Use: 1/f = 1/v − 1/u; P = 1/f
1/f = 1/(−100) − 1/(−25) = −1/100 + 4/100 = 3/100
f = 100/3 = 33.3 cm = 0.333 m
P = 1 / 0.333 = +3 D
Answer: +3 D (convex lens)
A person's near point is 50 cm. Find the power of the spectacle lens.
Given: u = −25 cm; v = −50 cm
Use: 1/f = 1/v − 1/u
1/f = −1/50 + 1/25 = 1/50
f = 50 cm = 0.5 m, so P = +2 D
Answer: +2 D
A hypermetropic person has a near point of 2 m. Find the power of the lens.
Given: u = −25 cm; v = −200 cm
Use: 1/f = 1/v − 1/u
1/f = −1/200 + 1/25 = (−1 + 8)/200 = 7/200
f = 200/7 = 28.6 cm = 0.286 m
P = 7/2 = +3.5 D
Answer: +3.5 D
Find the power of the lens for a person whose near point is 40 cm.
Given: u = −25 cm; v = −40 cm
Use: 1/f = 1/v − 1/u
1/f = −1/40 + 1/25 = (−5 + 8)/200 = 3/200
f = 200/3 = 66.7 cm = 0.667 m
P = +1.5 D
Answer: +1.5 D
A person uses a +2.5 D lens for reading at 25 cm. Where is the person's unaided near point?
Given: P = +2.5 D, so f = +40 cm; u = −25 cm
Use: 1/v = 1/f + 1/u
1/v = 1/40 + 1/(−25) = (5 − 8)/200 = −3/200
v = −66.7 cm
The image is virtual at 66.7 cm, which is the near point.
Answer: About 66.7 cm
Because of atmospheric refraction the Sun is seen 2 minutes before actual sunrise and 2 minutes after actual sunset. How much extra daylight is this per day and over 30 days?
Given: Gain = 2 min + 2 min per day
Use: Total = daily gain × days
Per day: 2 + 2 = 4 min
In 30 days: 4 × 30 = 120 min = 2 hours
Answer: 4 minutes a day; 2 hours in 30 days
Blue light has a wavelength of 400 nm. Red light has a wavelength about 1.8 times as long. Find the wavelength of red light.
Given: λ_blue = 400 nm; ratio = 1.8
Use: λ_red = 1.8 × λ_blue
λ_red = 1.8 × 400 = 720 nm
Answer: About 720 nm
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The ability of the eye lens to adjust its focal length, by changing its curvature with the help of the ciliary muscles, so that both near and distant objects are focused clearly on the retina.
A myopic eye needs a concave lens. Its focal length equals the far point, f = −1.2 m, so the power is P = 1/f = −0.83 D.
Answer: Concave lens of about −0.83 D
The near point is about 25 cm from the eye, and the far point is at infinity.
The student is suffering from myopia (near-sightedness). It can be corrected by using spectacles with a concave lens of suitable power.
Adjusting the focal length of the eye lens is called accommodation.
Answer: (b)
Answer: (d) retina
Answer: (c) 25 cm
Answer: (c) ciliary muscles
(i) f = 1/P = 1/(−5.5) = −0.18 m = −18.2 cm (a concave lens).
(ii) f = 1/P = 1/(+1.5) = +0.67 m = +66.7 cm (a convex lens).
Answer: −18.2 cm and +66.7 cm
The lens must form a virtual image of a very distant object at 80 cm. So u = ∞ and v = −80 cm, giving f = −80 cm = −0.8 m.
P = 1/f = −1.25 D.
Answer: Concave lens of power −1.25 D
The lens must make an object at 25 cm form a virtual image at 1 m. u = −25 cm, v = −100 cm.
1/f = 1/v − 1/u = −1/100 + 1/25 = 3/100, so f = 33.3 cm = 0.333 m.
P = 1/f = +3 D (a convex lens).
Answer: +3 D
To see an object closer than 25 cm, the ciliary muscles would have to make the eye lens thicker than they can. The focal length cannot be reduced below a certain minimum, so the image cannot be focused on the retina. The image is blurred and the eye feels strained.
The image distance stays the same. The retina is at a fixed distance from the lens. The eye adjusts the focal length of the lens (by the ciliary muscles) so that the image always forms on the retina.
Starlight is refracted continuously as it passes through layers of air of changing refractive index, so the apparent position of the star keeps changing slightly. The air is not steady, so the amount of light entering the eye also varies, and the star seems to flicker. A star is a point source, so this effect is clearly noticeable.
Planets are much closer to us and appear as extended sources, a collection of many point sources. The variations of light from the individual points average out, so the total amount of light entering the eye remains nearly constant and there is no twinkling.
At very great heights there is hardly any atmosphere, and so little scattering of sunlight. Without scattering no blue light reaches the astronaut's eyes from the sky, so it looks dark.