A container is filled with water (mu = 1.33 a | vedclass

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(C) The apparent depth of the object at the bottom of the container,as seen from above the water surface,is given by $d' = d / \mu$,where $d = 33.25 \

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) The apparent depth of the object at the bottom of the container,as seen from above the water surface,is given by $d' = d / \mu$,where $d = 33.25 \, cm$ and $\mu = 4/3$.$d' = 33.25 / (4/3) = 33.25 	imes 0.75 = 24.9375 \, cm \approx 25 \, cm$.The distance of the object from the mirror is $u = -(15 + 25) = -40 \, cm$.The image is formed $25 \, cm$ below the water level. The apparent depth of this image as seen from above the water surface is $v' = 25 / (4/3) = 25 	imes 0.75 = 18.75 \, cm$.The distance of the image from the mirror is $v = -(15 + 18.75) = -33.75 \, cm$.Using the mirror formula $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$:$\frac{1}{-33.75} + \frac{1}{-40} = \frac{1}{f}$$\frac{1}{f} = -(\frac{1}{33.75} + \frac{1}{40}) = -(\frac{40 + 33.75}{1350}) = -\frac{73.75}{1350}$$f = -\frac{1350}{73.75} \approx -18.3 \, cm$.Given the options,the closest value is $20 \, cm$ (assuming standard approximations in such problems).

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