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The refractive index of a lens material is 1.5 and focal length f. Due to some chemical changes in the material, its refractive index has increased by 2%. The percentage change in its focal length is

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+ 5.67%

– 5.67%

– 4.5%

+ 4.5%

answer is D.

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Detailed Solution

$\frac{1}{f} \propto (μ - 1), μ = 1.5$

$μ' = 1.5 + 1.5 [\frac{2}{100}]$

$\frac{f'}{f} = \frac{μ - 1}{μ' - 1}$

$\frac{f' - f}{f} \times 100 = \frac{- 0.03}{0.53} \times 10 = - 5.67 %$

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