Question

The near point of the eye of a person is 50 cm. Find the nature and power of the corrective lens required by the person to enable him to see clearly the objects placed at 25 cm from the eye?

Answer 1

Given:

Object distance (u) = −25cm

Image distance (v) = −50cm

Focal length, f = ?

Using the lens formula

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

`1/"f" = 1/(-50) - 1/(-25)`

`1/"f" = -1/50 + 1/25`

`1/"f" = (-1+2)/50`

`1/"f" = 1/50`

`"f"` = 50 cm

Since the focal length of the lens is positive, the corrective lens required is a convex lens. 

`"f"` = 50cm

`"f"` = 0.5m

Power (P) of the corrective lens will be 

`"P" = 1/("f"("m"))`

`"P" = 1/0.5`

`"P" = + 2 "D"`