Consider three media P , Q and R with refractive indices 1 , 1.25 ,...

The object $O$ is placed at the center of medium $Q$, which has a thickness of $5\text{ cm}$. Therefore, the actual depth of the object from both the $P-Q$ interface and the $Q-R$ interface is:

$d = \dfrac{5}{2} = 2.5\text{ cm}$

The formula for apparent depth when viewed from a medium of refractive index $n_{\text{observer}}$ into a medium of refractive index $n_{\text{object}}$ is given by:

$h_{\text{app}} = d \times \dfrac{n_{\text{observer}}}{n_{\text{object}}}$

When viewed from medium $P$ ($n_P = 1$), the apparent depth $h_1$ of the object in medium $Q$ ($n_Q = 1.25$) is:

$h_1 = 2.5 \times \dfrac{n_P}{n_Q} = 2.5 \times \dfrac{1}{1.25} = 2.5 \times \dfrac{4}{5} = 2\text{ cm}$

When viewed from medium $R$ ($n_R = 1.5$), the apparent depth $h_2$ of the object in medium $Q$ ($n_Q = 1.25$) is:

$h_2 = 2.5 \times \dfrac{n_R}{n_Q} = 2.5 \times \dfrac{1.5}{1.25} = 2.5 \times \dfrac{6/4}{5/4} = 2.5 \times \dfrac{6}{5} = 3\text{ cm}$

We need to find the value of $|h_1 - h_2|$:

$|h_1 - h_2| = |2 - 3| = |-1| = 1\text{ cm}$

Answer: $1$