Physics - Geometrical Optics Question with Solution | TestHub

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:

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

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:

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

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

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

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:

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

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

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

Answer: $1$