An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus of film?

Options

A. 5.6 m
B. 7.2 m
C. 2.4 m
D. 3.2 m

Answer

D

Explanation

`u=-240cm`
`v=12cm`
From lens formula
`(1)/(v)-(1)/(u)=(1)/(f)`
`f=(uv)/(u-v)=(-240xx12)/(-240-12)=(80)/(7)cm`

Normal shift by glass `=t(1-(1)/(mu))=(1-(2)/(3))=(1)/(3)cm`
Now, the image distance should be as :
`v.=v-(1)/(3)=12-(1)/(3)=(35)/(3)cm`
From lens formula
`-(1)/(u.)=(1)/(f)-(1)/(v.)=(7)/(80)-(3)/(35)`
`=(49-48)/(560)`
`u.=-560cm`
`=-5.6m`
Shift in position of object `=5.6-2.4=3.2 m`