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(B) Light incident from particle $P$ will be reflected at the mirror.For the mirror,the object distance $u = -5\, cm$ and the focal length $f = -R/2 =

Vedclass · Accepted Answer

$8.8$

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(B) Light incident from particle $P$ will be reflected at the mirror.For the mirror,the object distance $u = -5\, cm$ and the focal length $f = -R/2 = -20\, cm$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:$\frac{1}{v} + \frac{1}{-5} = \frac{1}{-20}$$\frac{1}{v} = \frac{1}{5} - \frac{1}{20} = \frac{4-1}{20} = \frac{3}{20}$$v = +\frac{20}{3}\, cm$.This image acts as a virtual object for the light getting refracted at the water surface.The distance of this virtual object from the water surface is $d_{obj} = 5 + \frac{20}{3} = \frac{35}{3}\, cm$.Since the light travels from water $(\mu_1 = 4/3)$ to air $(\mu_2 = 1)$,the apparent depth $d'$ is given by $d' = d_{obj} 	imes (\mu_2 / \mu_1)$.$d' = \left( \frac{35}{3} ight) 	imes \left( \frac{1}{4/3} ight) = \frac{35}{3} 	imes \frac{3}{4} = \frac{35}{4} = 8.75\, cm$.Thus,$d \approx 8.8\, cm$.

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