A lens of refractive index n is placed in a l | vedclass

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(A) The focal length of a lens in air is given by the Lens Maker's Formula: $\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.When

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$ - \frac{fn'(n - 1)}{n' - n}$

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(A) The focal length of a lens in air is given by the Lens Maker's Formula: $\frac{1}{f} = (n - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.When the lens is placed in a liquid of refractive index $n'$,the focal length $f'$ is given by: $\frac{1}{f'} = \left( \frac{n}{n'} - 1 \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) = \left( \frac{n - n'}{n'} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$.Dividing the two equations: $\frac{f'}{f} = \frac{(n - 1)}{\frac{n - n'}{n'}} = \frac{n'(n - 1)}{n - n'}$.Rearranging for $f'$: $f' = \frac{fn'(n - 1)}{n - n'} = - \frac{fn'(n - 1)}{n' - n}$.

$A$ lens of refractive index $n$ is placed in a liquid of refractive index $n'$. If the focal length of the lens in air is $f$,what will be its focal length in the liquid?

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