Question

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 \[ |h_1-h_2|. \]

• A. 3 cm
• B. 0 cm
• C. 1 cm
• D. 2 cm

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.

Answer 1

The Correct Option is C

Concept: For observation near the normal, \[ \text{Apparent depth} = \text{Real depth} \times \frac{n_{\text{observer}}}{n_{\text{object medium}}} \]

Step 1: Locate the object
Thickness of medium \(Q\) \[ =5\,\text{cm} \] Object is at the centre. Hence distance from either surface \[ =2.5\,\text{cm} \]

Step 2: Find apparent depth when viewed from medium P
Observer is in medium \(P\), \[ n_P=1 \] Object is in medium \(Q\), \[ n_Q=1.25 \] Thus, \[ h_1 = 2.5 \left( \frac{1}{1.25} \right) \] \[ h_1=2\,\text{cm} \]

Step 3: Find apparent depth when viewed from medium R
Observer is in medium \(R\), \[ n_R=1.5 \] Therefore, \[ h_2 = 2.5 \left( \frac{1.5}{1.25} \right) \] \[ h_2=3\,\text{cm} \]

Step 4: Calculate the difference
\[ |h_1-h_2| = |2-3| \] \[ |h_1-h_2| = 1\,\text{cm} \] \[ \boxed{1\,\text{cm}} \] Hence, \[ \boxed{\text{Option (C)}} \]