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. 50 cm towards the lens
B. 50 cm away from the lens
C. 12.5 cm towards the lens
D. 12.5 cm away from the lens

Answer

D

Explanation

According to the question

Given focal length `f=25 cm`
and distance between image of an object and screen
Now By lens formula
`(1)/(f ) = (1)/( v ) - (1)/(u)`
[`:.` Because screen is moved closer to teh lens ]
`(1)/(u ) = (1)/(f ) - (1)/(v )`
`rArr = u= (75 xx 25 )/(50) = (75)/(2) cm`
when the screen shift upto by 25 cm then the screen will be at 2f
For to get sharp image object has to be at 2f
So the distance is v- u = f
`50- (75)/(2) =(25)/(2) = 12.5 cm`
This is the distance through which the object shifted from the lens.