Physics - Geometrical Optics Question with Solution | TestHub

From the figure, object O is at height 2h/3 above the mirror at the base.

The plane mirror forms the image as far behind it as the object is in front:$\text{image is }\frac{2h}{3}\text{ behind the mirror}$

$\text{Real depth of image below the liquid surface}=h+\frac{2h}{3}=\frac{5h}{3}$

$\text{Apparent depth}=\frac{\text{real depth}}{\mu }=\frac{5h}{3\mu }$

$\text{Distance of image from observer E}=H+\frac{5h}{3\mu }$

Hence option (B).