# Human Eye and Colourful World

## NCERT Exercise Questions

### Part 1

Question 1: The human eye can focus objects at different distances by adjusting the focal length of the eye lens. This is due to

- Presbyopia
- Accommodation
- Near sightedness
- Far sightedness
**Answer:**(b) Accommodation

Question 2: The human eye forms the image of an object at its

- Cornea
- Iris
- Pupil
- Retina
**Answer:**(d) Retina

Question 3: The least distance of distinct vision for a young adult with normal vision is about

- 25 m
- 2.5 cm
- 25 cm
- 2.5 m
**Answer:**(c) 25 cm

Question 4: The change in focal length of an eye lens is caused by the action of the

- Pupil
- Retina
- Ciliary muscles
- Iris
**Answer:**(c) Ciliary muscles

Question 5: 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

Question 6: 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

Question 7: 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.