A man can see the object between `15 cm` and `30 cm`. He uses the lens to see the far objects. Then due to the lens to see the far objects. Then due to the lens used, the near point will be at

Options

A. `(10)/(3)cm`
B. `30 cm`
C. `15 cm`
D. `(100)/(3) cm`

Answer

B

Explanation

For improving far oint, concave lens if requuired and for this concave lens`u=oo,v=-30cm`
So `(1)/(f)=(1)/(-30)-(1)/(oo)impliesf=-30cm`
for near point `(1)/(-30)=(1)/(-15)-(1)/(-15)impliesu=-30cm`