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One surface of a biconvex lens having focal length 40 cm is silvered. The radius of the curvature of the other surface is 60 cm. At what distance the silvered lens should an object be placed to obtain a real image magnified three times ?
(Refractive index of lens, μL=1.5)

A
10 cm
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B
11.4 cm
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C
12 cm
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D
12.4 cm
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Solution

The correct option is B 11.4 cm
Given:
fL=40 cm; R1=60 cm


Let, x= magnitude of radius of the curvature of the silvered surface.

From lens maker's formula:

1fL=(μLμm1)(1R11R2)

where, μLRI of lens =1.5
μmRI of medium =1

140=(1.511)(1+601x)

140=12(160+1x)

x=60×206020=30 cm

The silvered lens will behave like a converging mirror with focal length Fe because it is a combination of a converging lens and a converging mirror.

1Fe=1fL+1fm+1fL .....(1)

where, fL=40 cm
fm = focal length of mirror=302=15 cm

Putting the value of above data in Eq.(1)

1Fe=240115Fe=607 cm

Focal length of the mirror is negative means the mirror is concave mirror.


For the concave mirror
Magnification =3, Fe=607cm

M=ffu

3=607607u

3u+1807=607

u=807 cm=11.42 cm
Hence, option (b) is correct.

Why this Question?CAUTION: Entire setup acts like a mirror so the magnificationformula of the mirror is to be used

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