Question

Doctor tested the eyes of Rajkumar and identified that he had hypermetropla. The distance of near point ls 50 cm. The power of the lens suggested by the doctor to him is

A)
- 2D

B)
+ 1D

C)
- 1D

D)
+ 2D​

Answer 1

Explanation:

The person needs a convex lens to rectify this defect.

Calculation of power of the lens:

This hypermetropic eye can see the nearby object kept at 25 cm clearly if the image is formed at its own near point i.e. 50 cm.

Object distance, u=−25 cm

Image distance, v=−50 cm

v

1

−

u

1

=

f

1

−50

1

−

−25

1

=

f

1

f=50 cm

P=

f

100

=

50

100

=2D