A convex lens is immersed in water. How will | vedclass

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(B) The power of a lens is given by the formula $P = \frac{1}{f} = (n_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,where $n_l$ is the relative

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It will decrease.

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(B) The power of a lens is given by the formula $P = \frac{1}{f} = (n_l - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$,where $n_l$ is the relative refractive index of the lens material with respect to the surrounding medium.When the lens is in air,$n_l = \frac{n_g}{n_a} \approx n_g$. Since $n_g > 1$,the lens is converging.When the lens is immersed in water,the relative refractive index becomes $n_l' = \frac{n_g}{n_w}$.Since the refractive index of water $(n_w \approx 1.33)$ is greater than that of air $(n_a \approx 1.0)$,the relative refractive index $n_l'$ decreases $(n_l' As $P \propto (n_l - 1)$,a decrease in the relative refractive index leads to a decrease in the power $P$ of the lens.

$A$ convex lens is immersed in water. How will its power $P$ change?

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