A concave mirror and a converging lens (glass | vedclass

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(A) $1$. The focal length of a mirror depends only on its radius of curvature $(f = R/2)$ and is independent of the surrounding medium. Therefore,the

Vedclass · Accepted Answer

$f_{Lens} = 12 \, cm, f_{Mirror} = 3 \, cm$

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(A) $1$. The focal length of a mirror depends only on its radius of curvature $(f = R/2)$ and is independent of the surrounding medium. Therefore,the focal length of the concave mirror remains $3 \, cm$ in water.$2$. The focal length of a lens is given by the Lens Maker's Formula: $\frac{1}{f} = (\mu_{rel} - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} ight)$,where $\mu_{rel} = \frac{\mu_{lens}}{\mu_{medium}}$.$3$. In air: $\frac{1}{f_a} = (1.5 - 1) K = 0.5 K$,where $K = (\frac{1}{R_1} - \frac{1}{R_2})$. Given $f_a = 3 \, cm$,so $K = \frac{1}{1.5} = \frac{2}{3}$.$4$. In water: $\mu_{rel} = \frac{1.5}{4/3} = 1.5 	imes \frac{3}{4} = 1.125 = \frac{9}{8}$.$5$. $\frac{1}{f_w} = (\frac{9}{8} - 1) K = (\frac{1}{8}) 	imes \frac{2}{3} = \frac{1}{12}$.$6$. Thus,$f_w = 12 \, cm$.

$A$ concave mirror and a converging lens (glass with $\mu = 1.5$) both have a focal length of $3 \, cm$ when in air. When they are in water $\left( \mu = \frac{4}{3} \right)$,what are their new focal lengths?

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Assertion : $A$ concave mirror and convex lens both have the same focal length in air. When they are submerged in water,they will have same focal length.
Reason : The refractive index of water is smaller than the refractive index of air.

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