Consider the situation shown in the figure. W | vedclass

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(B) The apparent depth of the object $O$ at the bottom of the beaker is given by $d' = \frac{d}{\mu_w}$,where $d = 10 \, cm$ is the real depth and $\m

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$12.5$

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(B) The apparent depth of the object $O$ at the bottom of the beaker is given by $d' = \frac{d}{\mu_w}$,where $d = 10 \, cm$ is the real depth and $\mu_w = \frac{4}{3}$ is the refractive index of water.$d' = \frac{10}{4/3} = \frac{30}{4} = 7.5 \, cm$.This means the object $O$ appears to be at a position $O'$ at a depth of $7.5 \, cm$ below the water surface.The distance of this apparent position $O'$ from the plane mirror is the sum of the height of the mirror above the water surface and the apparent depth of the object.Distance $= 5 \, cm + 7.5 \, cm = 12.5 \, cm$.Since a plane mirror forms an image at the same distance behind it as the object is in front of it,the final image will be formed at $12.5 \, cm$ behind the mirror.

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