• question_answer A lens of refractive index n is put in a liquid of refractive Index n' of focal length of lens in air is f, its focal length in liquid will be :- A)  $-\frac{f\,n'(n-1)}{n'-n}$B)  $-\frac{f(\,n'-n)}{n'(n-1)}$C)  $-\frac{n'(\,n-1)}{f\,(n'-n)}$D)  $\frac{fn'n}{n-n'}$

Solution :

$\frac{1}{f}=\left( \frac{n-1}{1} \right)\,\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)$and $\frac{1}{f'}=\left( \frac{n-n'}{n'} \right)\,\left( \frac{1}{{{R}_{1}}}-\frac{1}{{{R}_{2}}} \right)$ $\therefore$     $\frac{f'}{f}=\frac{n-1}{1}\times \frac{n'}{n-n'}$ $\Rightarrow$$f=\frac{fn'(n-1)}{n'-n}$

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