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Question

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 from the silvered lens should an object be placed to obtain a real image magnified three times? (μg=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 that,
Focal length of biconvex lens, fl=40 cm
Radius of curvature of non silvered lens, R1=60 cm
Refractive index of lens, μg=1.5

Let x is the magnitude of radius of the curvature of the silvered surface.
Sign convention: Direction along the incident ray is taken as positive.


From lens maker's formula:
1fl=(μgμ01)(1R11R2)
140=(1.511)(1+601x)
140=12(160+1x)
x=60×206020=30 cm
x=R2=30 cm
Now
fm= focal length of mirror
fm=R2=15 cm

The silvered lens will behave like a converging mirror with focal length F because it is a combination of a converging lens and a converging mirror
1F=2fl+1fm


Where fl=40 cm
1F=240+115
F=607 cm
For the concave mirror
Magnification =3, F=60/7 cm
M=fuf
3=+60/7u+60/7
3u3×607=607
3u=607+1807
3u=2407
u=807 cm
u=11.42 cm
Why this question ?
Note: Entire setup acts like a mirror so the magnification formula of the mirror is to be used.


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