A thin converging lens of focal length f= 25 cm forms the image of an object on a screen placed at a distance of 75 cm from the lens. The screen is moved closer to the lens by a distance of 25 cm. The distance through which the object has to be shifted, so that its image on the screen in sharp again is

Options

A. 37.5 cm
B. 16.25 cm
C. 12.5 cm
D. 13.5 cm

Answer

C

Explanation

According to first condition,
`1/f_(1)=1/v_(1) – 1/u_(1)``implies u_(1)=(-75)/2 += -37.5`
According to second condition,
`1/f_(2)=1/v_(2) – 1/u_(2)`
`1/25=1/50 – 1/u_(2)`
`therefore u_(2)= -50`
` therefore Delta u= u_(2)-u_(1)=50-37.5=12.5 cm`