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# If a lens of focal length, f, made of glass of refractive index 53 is placed in a liquid of refractive index, 4 53, then the focal length of the lens becomes2f−4f4ff

A
f
B
2f
C
4f
D
4f
Solution
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#### In Air 1f=(μg−1)[1R1+1R2]In water,1f′=(μgμw−1)[1R1+1R2]f′f=(μg−1μg−μw)×μwf′=⎛⎜ ⎜ ⎜⎝53−153−453⎞⎟ ⎟ ⎟⎠×453f=−−1718f

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