Class 10 Physics

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

• Presbyopia
• Accommodation
• Near sightedness
• Far sightedness

The human eye forms the image of an object at its

• Cornea
• Iris
• Pupil
• Retina

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

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

• Pupil
• Retina
• Ciliary muscles
• Iris

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.