Consider three media P, Q and R with refractive indices $1$, $1.25$, and $1.5$, respectively. The medium Q having a thickness of $5$ cm is placed between extended media P and R as shown in the figure. An object O is placed at the center of medium Q. If viewed from medium P near the normal direction, the apparent depth of O is $h_1$. For similar observation from medium R, the apparent depth is $h_2$. The value of $|h_1 - h_2|$, in cm, is :
Answer: (B) $1$
The object is at the centre of Q, so its real depth from either face is $2.5$ cm.
For near-normal viewing, apparent depth $= \text{real depth} \times \dfrac{n_{\text{observer}}}{n_{\text{object}}}$.
From P: $h_1 = 2.5 \times \dfrac{1}{1.25} = 2$ cm
From R: $h_2 = 2.5 \times \dfrac{1.5}{1.25} = 3$ cm
$$|h_1 - h_2| = 1\ \text{cm}$$
Solution by Sreeraj P, M.Sc Physics