Consider three media \(P\), \(Q\) and \(R\) with refractive indices
\[
n_P=1,\qquad n_Q=1.25,\qquad n_R=1.5
\]
respectively.
Medium \(Q\) has a thickness of \(5\,\text{cm}\) and is placed between media \(P\) and \(R\) as shown.
An object \(O\) is placed at the centre of medium \(Q\).
If viewed from medium \(P\) near the normal direction, the apparent depth of \(O\) is \(h_1\).
For the same object viewed from medium \(R\), the apparent depth is \(h_2\).
Find
\[
|h_1-h_2|.
\]
Show Hint
For normal viewing,
\[
\text{Apparent depth}
=
\text{Real depth}
\times
\frac{n_{\text{observer}}}{n_{\text{medium}}}
\]
Objects appear shallower when viewed from a rarer medium and deeper when viewed from a denser medium.
Step 1: Understand apparent depth. When we look at an object across a flat boundary near the normal, refraction makes it seem shifted. The rule is $h_{app} = h_{real}\times \dfrac{n_{observer}}{n_{object}}$, where the object sits in medium $Q$. Step 2: Find where the object actually is. Medium $Q$ is $5$ cm thick and the object $O$ is at its centre, so it lies $2.5$ cm below each surface. This $2.5$ cm is the real depth for both views. Step 3: View from medium P. Here the observer is in $P$ with $n_P=1$ and the object is in $Q$ with $n_Q=1.25$. So $h_1 = 2.5\times \dfrac{1}{1.25} = 2$ cm. Step 4: View from medium R. Now the observer is in $R$ with $n_R=1.5$, object still in $Q$. So $h_2 = 2.5\times \dfrac{1.5}{1.25} = 3$ cm. Step 5: Take the difference. \[ |h_1 - h_2| = |2 - 3| = 1 \text{ cm} \] Step 6: State the answer. The apparent depths differ by just $1$ cm, which matches option C. \[ \boxed{1 \text{ cm}} \]
