Human Eye and Colourful World class 10 science NCERT exercise solution

Human Eye

NCERT Solution

A person needs a lens of power –5.5 dioptres for correcting his distant vision. For correcting his near vision he needs a lens of power +1.5 dioptre. What is the focal length of the lens required for correcting (i) distant vision, and (ii) near vision?

Answer: The power of a lens can be calculated by following formula:

`P = 1/f`

In case of distant vision, the given power is -5.5 dioptre.

Or, `-5.5 = 1/f`

Or, `f = 1 ÷ -5.5 = -0.18  m`

Hence, the focal length of lens required for distant vision = -1.8 cm

In case of near vision, the given power is +1.5 D

Or, `1.5 D = 1/f`

Or, f = `1 ÷ 1.5 = 0.67  m`

Hence, the focal length of the required lens for near vision = +6.7 cm


The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to

  1. Presbyopia
  2. Accommodation
  3. Near sightedness
  4. Far sightedness

    Answer: (b) Accommodation

The human eye forms the image of an object at its

  1. Cornea
  2. Iris
  3. Pupil
  4. Retina

    Answer: (d) Retina

The least distance of distinct vision for a young adult with normal vision is about

  1. 25 m
  2. 2.5 cm
  3. 25 cm
  4. 2.5 m

    Answer: (c) 25 cm

The change in focal length of an eye lens is caused by the action of the

  1. Pupil
  2. Retina
  3. Ciliary muscles
  4. Iris

    Answer: (c) Ciliary muscles

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?

Answer: In the given situation, the person shall be able to see a distant object clearly if the image of that object would be formed at his far point, which is 80 cm.

Given; object distance u = ∞

Image distance v = -80 cm = - 0.8 m

The focal length and power of the required lens can be calculated by using the following formula:

`1.v-1/u=1/f`

Or, `-(1)/(0.8 m)-0=1/f`

Or, `1/f=-(1)/(0.8)`

Or, `P=-1.25 D`

The negative sign shows that the lens is a concave lens.

The power of lens = -1.25 D


Make a diagram to show how hypermetropia is corrected. The near point of a hypermetropic eye is 1 m. What is the power of the lens required to correct this defect? Assume that the near point of the normal eye is 25 cm.

Answer: In the given situation, the person shall be able to see a clear image, if the image of an object kept at 25 cm would be formed at the near point of this person which is 1 m.

Given, object distance u = -25 cm = - 0.25 m

Image distance v = - 1 m

The focal length and power of the required lens can be calculated using following formula:

`1/v-1/u=1/f`

Or, `-(1)/(1 m)=-(1)/(-0.25 m)=1/f`

Or, `P=-1+4=3 D`

The positive sign shows that it is a convex lens.



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