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Question

# A lens of refractive index μ is put in a liquid of refractive index μ′. If the focal length of lens in air is f, then its focal length in liquid will be

A
fμ(μ1)(μμ)
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B

f(μμ)μ(μ1)

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C

μ(μ1)f(μμ)

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D
fμμ(μμ)
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Solution

## The correct option is A −fμ′(μ−1)(μ′−μ)According to question. here, μ = refractive index of lens, μ' = refractive index of liquid. f= focal length of lens in air. Case I: When lens is in air, From the lens maker's formula, 1f=(μ−1)(1R1−1R2),...(i) Case II: When lens is dipped in liquid. 1f′=(μμ′−1)(1R1−1R2) =(μ−μ′μ′)(1R1−1R2)...(ii) From equation (i) and (ii), we get ∴f′f=(μ−1)μ′(μ−μ′)=−(μ−1)μ′(μ′−μ) f′=−fμ′(μ−1)(μ′−μ)

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