CBSE Class 10 Science The Human Eye and the Colourful World Sure Shot Questions Set 03

This Sure Shot page gathers 14 of the most expected board-style questions for CBSE Class 10 Science Chapter 10, The Human Eye and the Colourful World, with full step-wise answers and redrawn ray diagrams. It covers accommodation, the parts of the eye, myopia and hypermetropia with ray diagrams, bifocal lenses, lens-power numericals, refraction and dispersion through a prism, cataract, presbyopia and the rainbow.

Most Expected Questions

Question 1. Define the term power of accommodation. Write the modification in the curvature of the eye lens which enables us to see the nearby objects clearly.
Answer:
The ability of the eye lens to adjust its focal length is called the power of accommodation. It is done by the ciliary muscles, which change the curvature of the lens.
To see nearby objects, the ciliary muscles contract, the curvature of the eye lens increases (the lens becomes thicker) and its focal length decreases. This lets the image of the nearby object fall on the retina.
In simple words: The eye lens becomes thicker for near objects.

Teacher's Note:
1. Define accommodation and also state the change.
2. Focal length decreases when curvature increases.

Question 2. Write about the power of accommodation of the human eye. Explain why the image distance in the eye does not change when we change the distance of an object from the eye.
Answer:
The ability of the eye lens to adjust its focal length by the action of the ciliary muscles is called the power of accommodation.
The image must always be formed on the retina, which is at a fixed distance from the eye lens. When the object distance changes, the ciliary muscles change the curvature and hence the focal length of the lens: for a distant object the muscles relax, the lens becomes thinner and its focal length increases; for a nearby object they contract, the lens becomes thicker and its focal length decreases.
So the focal length changes, and the image distance stays the same (about 2.5 cm).
In simple words: The retina is fixed, so the lens changes its focal length instead.

Teacher's Note:
1. Mention both near and far cases.
2. Do not say the retina moves.

Question 3. Trace the sequence of events which occur when a bright light is focused on your eyes.
Answer:
1. Most of the refraction of the light entering the eye takes place at the outer surface of the cornea.
2. The pupil, controlled by the iris, becomes smaller to reduce the amount of light entering the eye in bright light.
3. The crystalline lens provides the finer adjustment of the focal length and forms the image on the retina.
4. The light-sensitive cells of the retina get activated and generate electrical signals.
5. The signals are sent to the brain through the optic nerves, and the brain interprets them so that we perceive the object.
In simple words: Light is bent by the cornea, adjusted by the pupil and lens, sensed by the retina and understood by the brain.

Teacher's Note:
1. Write the events in order.
2. Name the part for each step.

Question 4. (a) List the parts of the human eye that control the amount of light entering into it. Explain how they perform this function. (b) Write the function of the retina in the human eye.
Answer:
(a) The iris and the pupil control the amount of light. The iris is a muscular diaphragm and the pupil is the opening in its centre. In bright light the iris contracts and the pupil becomes small, so less light enters. In dim light the iris relaxes and the pupil becomes large, so more light enters.
(b) The retina is the light-sensitive screen on which the image is formed. Its rod and cone cells convert it into electrical signals, which are sent to the brain through the optic nerve.
In simple words: The iris changes the pupil size, and the retina turns the image into signals.

Teacher's Note:
1. Describe bright and dim light separately.
2. Retina has rods and cones.

Question 5. (a) A person is suffering from both myopia and hypermetropia. (i) What kind of lenses can correct this defect? (ii) How are these lenses prepared? (b) A person needs a lens of power +3 D for correcting his near vision and −3 D for correcting his distant vision. Calculate the focal lengths of the lenses required to correct these defects.
Answer:
(a)(i) A bifocal lens.
(a)(ii) It is prepared with a concave lens in the upper portion, for distant vision, and a convex lens in the lower portion, for near vision.
(b) Near vision: P = +3 D, so f = 1/P = 1/3 = 0.33 m = +33.3 cm (convex lens).
Distant vision: P = −3 D, so f = 1/(−3) = −0.33 m = −33.3 cm (concave lens).
In simple words: Bifocal lenses have two parts: concave for far and convex for near.

Teacher's Note:
1. f = 1/P in metres.
2. Sign shows the lens type.

Question 6. What is myopia? List two causes for the development of this defect. How can this defect be corrected using a lens? Draw ray diagrams to show the image formation in case of (i) the defective eye and (ii) the corrected eye.
Answer:
Myopia (short-sightedness) is a defect of vision in which a person can see nearby objects clearly but cannot see distant objects clearly. The image of a distant object is formed in front of the retina.
Causes: (i) the eyeball is too long (elongated); (ii) the eye lens is too curved, so its focal length is too short.
Correction: by a concave lens of suitable power, which diverges the rays a little before they enter the eye, so the image is formed on the retina.

In simple words: A myopic eye focuses in front of the retina, and a concave lens moves the image back.

Teacher's Note:
1. Show the image in front of the retina in the first diagram.
2. Show it exactly on the retina in the second.

Question 7. (a) List two causes of hypermetropia. (b) Draw ray diagrams showing (i) a hypermetropic eye and (ii) its correction using a suitable optical device.
Answer:
(a) Causes: (i) the eyeball is too short; (ii) the focal length of the eye lens is too long (its converging power is too low).
(b) In (i) the image of a nearby object is formed behind the retina. In (ii) a convex lens placed in front of the eye converges the rays slightly, so the image is formed on the retina.

In simple words: A hypermetropic eye focuses behind the retina, and a convex lens brings the image forward.

Teacher's Note:
1. Show the image behind the retina.
2. Use a convex lens for correction.

Question 8. While sitting in the last row, a student has difficulty in reading the blackboard clearly. State the defect of vision the student is suffering from. Mention two causes of this defect. Suggest a suitable lens for the correction of this defect.
Answer:
The student is suffering from myopia (short-sightedness).
Causes: (a) the distance of the retina from the eye lens has increased (the eyeball is elongated); (b) the focal length of the eye lens has become too small.
Correction: a concave lens of suitable focal length (power).
In simple words: The student is short-sighted and needs concave spectacles.

Teacher's Note:
1. Give the defect, the causes and the lens in that order.
2. Concave lenses diverge the rays.

Question 9. A student suffering from myopia is not able to see distinctly the objects placed beyond 5 m. (a) List two possible reasons due to which this defect of vision may have arisen. With the help of ray diagrams, explain (i) why the student is unable to see distinctly the objects placed beyond 5 m from his eyes, (ii) the type of the corrective lens used to restore proper vision and how this defect is corrected by the use of this lens. (b) If, in this case, the numerical value of the focal length of the corrective lens is 5 m, find the power of the lens as per the new Cartesian sign convention.
Answer:
(a) Reasons: the eyeball is too long, or the eye lens is too curved (focal length too short).
(i) The far point of his eye is 5 m. Rays from objects beyond 5 m are focused in front of the retina, so the image is blurred (see diagram (a)).
(ii) A concave lens is used. It diverges the rays so that they seem to come from the far point (5 m). The eye lens can then focus them on the retina (see diagram (b)).

(b) The lens is concave, so by the new Cartesian sign convention f = −5 m. P = 1/f = 1/(−5) = −0.2 D.
In simple words: A concave lens of −0.2 D lets him see beyond 5 m.

Teacher's Note:
1. The numerical value of f is 5 m; the sign is negative for a concave lens.
2. The far point and the focal length have the same magnitude.

Question 10. Draw a ray diagram to show the refraction of light through a glass prism. Mark on it (a) the incident ray, (b) the emergent ray and (c) the angle of deviation.
Answer:
In the diagram the incident ray falls on face AB and bends towards the normal; inside the prism it travels to face AC and bends away from the normal on emerging. The emergent ray is bent towards the base BC of the prism. ∠i is the angle of incidence, ∠e the angle of emergence, ∠A the angle of the prism and ∠D the angle of deviation, which is the angle between the incident ray produced forward and the emergent ray produced backward.

In simple words: A prism bends the ray towards its base; the angle between the original and final directions is the angle of deviation.

Teacher's Note:
1. Label the incident ray, the emergent ray and ∠D.
2. Draw the normals at both faces.

Question 11. What is 'dispersion of white light'? State its cause. Draw a ray diagram to show the dispersion of white light by a glass prism.
Answer:
Dispersion is the splitting of white light into its seven constituent colours (violet, indigo, blue, green, yellow, orange, red) when it passes through a glass prism. The band of colours is the spectrum.
Cause: the colours of white light travel with different speeds in glass, so the refractive index of glass is different for each colour (largest for violet, smallest for red). Each colour is therefore bent through a different angle: red the least and violet the most.

In simple words: Glass bends each colour by a different amount, so white light spreads into a spectrum.

Teacher's Note:
1. Mark red at the top and violet at the bottom on the screen.
2. Say that the colours separate because their speeds differ in glass.

Question 12. Why do we see a rainbow in the sky only after rainfall?
Answer:
A rainbow is formed by the dispersion of sunlight by tiny water droplets suspended in the air. After rainfall the air contains many such droplets.
Each droplet acts like a small prism: it refracts and disperses the sunlight, reflects it internally, and refracts it again as it comes out. Different colours leave at different angles and reach the observer's eye as a rainbow.
The Sun must be behind the observer for this to be seen.
In simple words: Raindrops left in the air split the sunlight into colours.

Teacher's Note:
1. Droplets are needed for a rainbow.
2. Sun behind the observer.

Question 13. What are the causes of the following defects of vision and how can they be corrected? (a) Cataract (b) Presbyopia.
Answer:
(a) Cataract: in old age the crystalline lens sometimes becomes milky and cloudy, causing partial or complete loss of eyesight. It is corrected by cataract surgery.
(b) Presbyopia: the power of accommodation of the eye decreases with age because the ciliary muscles weaken and the eye lens becomes less flexible, so the near point moves away. It is corrected by spectacles with convex lenses, or bifocal lenses if myopia is also present.
In simple words: Cataract is a cloudy lens treated by surgery; presbyopia is weak focusing corrected by bifocals.

Teacher's Note:
1. Give the cause and the correction for both.
2. Cataract cannot be corrected by spectacles.

Question 14. A person needs a lens of power −4.5 D for correction of his/her vision. (A) What kind of defect in vision is she suffering from? (B) What is the focal length of the corrective lens? (C) What is the nature of the corrective lens?
Answer:
(A) The power is negative, so the defect is myopia (short-sightedness).
(B) f = 1/P = 1/(−4.5) = −0.222 m = −22.2 cm.
(C) The lens is a concave (diverging) lens.
In simple words: A −4.5 D lens is a concave lens of focal length −22.2 cm for myopia.

Teacher's Note:
1. Negative power: concave lens, myopia.
2. Convert to cm for the focal length.