
Here we are providing numerical assignment on class 10 science chapter human eye and colourful world. Answers are also provided for reference.
Numerical Problems Based on Human Eye and Colourful World
Q.1. If the concave lens of focal length (f = 1.5m) used to restore the proper vision, then what is the power of lens.
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Answer:
Q.2. A young boy can adjust the power of his eye-lens between 50 D & 60 D. His far point is infinity.
(a) What is the distance of his retina from the eye-lens?
(b) What is his near point?
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Answer: (a) When the eye is fully relaxed, its focal length is largest and the power of the eye lens is minimum.This power is 50 D according to the given data. The focal length is 1/50 m = 2cm. As the far point is at infinity, the parallel rays coming from infinity are focused on the retina in the fully relaxed condition. Hence, the distance of the retina from the lens equal to the focal length which is 2 cm.
(b)

Q.3. A person cannot see objects clearly beyond 50 cm. Find the power of the lens to correct the
vision.
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Answer:
Q.4. A myopic person having far point 80 cm uses spectacles of power –1.0 D. How far can he see
clearly?
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Answer:
Q.5. A person having a myopic eye used the concave lens of focal length 50 cm. What is the power of the lens.
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Answer:
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Q.6. A 52-year-old near-sighted person wears eye glass of power of –5.5D for distance viewing. His
doctor prescribes a correction of +1.5D in the near-vision section of his bi-focals this measured
relative to the main parts of the lens
(i) What is the focal length of his distance viewing part of the lens.
(ii) What is the focal length of the near vision section of the lens.
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Answer:
Q.7. A short-sighted person cannot see clearly beyond 2 m. Calculate the power of less required to correct his vision.
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Answer:
Q.8. A person is able to see objects clearly only when these are lying at distance between 50 cm and 300 cm from his eye.
(i) What kind of defect of vision he is suffering from ?
(ii) What kind of lenses will be required to increase his range of vision from 25 cm to infinity? Explain briefly.
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Answer: (a) For a normal eye, the near point is at 25 cm and the far point is at infinity from the eye. The given person cannot see object clearly either closer to the eye or far away from the eye. So, he is suffering from both myopia and hypermetropia.(b) A bi-focal lens consisting of a concave lens and convex lens of suitable focal lengths will be required to correct the defects and to increase his range of vision form 25 cm to infinity. In a bi-focal lens, upper portion is concave which corrects distant vision and lower portion is convex which corrects near vision.
Q.9. A 14-years old student is not able to see clearly to question written on a black board placed at a distance of 5 m from him.
(a) Name the defect of vision he is suffering from.
(b) Name the type of lens used to correct this defect.
(c) State two causes of this defect.
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Answer: (a) Student is suffering from Myopia or short-sightedness(b) Concave lens of suitable focal length is used to correct this defect.
(c) It is due to (i) elongation of the eye ball, (ii) excessive curvature of the cornea.
Q.10. A person suffering from far – sightedness wears a spectacle having a convex lens of focal length 50 cm. What is the distance of the near point of his eye?
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Answer: 50 cmNumerical Assignment (with Answers)
1. A person with a myopic eye cannot see objects beyond 1.2 m distinctly. What should be the nature of corrective lens to restore proper vision? [Ans: P= -0.83 Lens – concave]
2. The near point of a hypermetropic eye is at 75 cm from the eye. What is the power of the lens required to enable him to read clearly a book held at 25 cm from the eye? [Ans: P= 2.66D]
3. A myopic person uses specs of power – 0.5 D. What is the distance of far point of his eye? (Ans: 2 m)
4. A person wants to read a book placed at 20 cm, whereas near point of his eye is 30 cm. calculate the power of the lens required. [Ans: 1.67 D]
5. The far point distance of a short sighted person is 1.5 meters. find the focal length, power and nature of the remedial lens? (Ans: -1.5 m, -0.67 D, concave lens)
6. A person having a myopic eye uses a concave lens of focal length 10 cm. Find the power of the lens. (Ans: -10 D)
7. A person with myopic eye cannot see objects beyond 1.2 m distinctly. What should be the nature of corrective lenses to restore proper vision?
Ans: Here, distance of far point, x = 1.2 m
For viewing distant objects, focal length of corrective lens, F = -x = -1.2 m,
P = 1 / f = 1 / -1.2 = -0.83D
8. The far point of a myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to correct the problem?
Ans: For diagram, Here, x’ = 1 m = 100 cm, d = 25 cm, f =?
From f = x’d / x’- d, F = 100*25 / 100 – 25 = 33.3 cm, P = 100 / f = 100 / 33.3 = 3 D
9. The far point of myopic person is 80 cm in front of the eye. What is the nature and power of the lens required to enable him to see very distant objects distinctly? [Ans: Concave; -1.25 D]
10. The far point of a myopic person is 150 cm in front the eye. Calculate the focal length and power of a lens required to enable him to see distant objects clearly. [Ans: -1.5 m, -0.67 D]

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