Consider the situation shown in the figure. W | vedclass

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(B) $1$. First,calculate the apparent depth of the object $O$ as seen from the air surface. The formula for apparent depth is $d_{app} = \frac{d_{real

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$12.5$

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(B) $1$. First,calculate the apparent depth of the object $O$ as seen from the air surface. The formula for apparent depth is $d_{app} = \frac{d_{real}}{\mu}$.$2$. Given $d_{real} = 10 \, cm$ and $\mu = 4/3$,the apparent depth is $d_{app} = \frac{10}{4/3} = \frac{30}{4} = 7.5 \, cm$.$3$. The object $O$ appears to be at a distance of $7.5 \, cm$ below the water surface.$4$. The distance of this apparent position of the object from the plane mirror is the sum of the distance from the mirror to the water surface $(5 \, cm)$ and the apparent depth $(7.5 \, cm)$.$5$. Total distance from the mirror = $5 \, cm + 7.5 \, cm = 12.5 \, cm$.$6$. Since a plane mirror forms an image at the same distance behind the mirror as the object is in front of it,the distance of the image from the mirror is $12.5 \, cm$.

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