Consider three media P, Q and R with refractive indices $1$, $1.25$, and $1.5$, respectively. The medium Q having a thickness of $5\text{ cm}$ 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 :
Given:
- Refractive index of medium P, nP = 1
- Refractive index of medium Q, nQ = 1.25
- Refractive index of medium R, nR = 1.5
- Thickness of medium Q = 5 cm
- Object O is at the center of medium Q.
Hence, the real depth of the object from either surface of Q is
Real depth = 5/2 = 2.5 cm
Step 1: Apparent depth when viewed from medium P
For refraction at a plane surface,
Apparent depth = Real depth × (Refractive index of observer's medium / Refractive index of object's medium)
Therefore,
h1 = 2.5 × (1/1.25) = 2 cm
Step 2: Apparent depth when viewed from medium R
Similarly,
h2 = 2.5 × (1.5/1.25) = 3 cm
Step 3: Find |h1 − h2|
|h1 − h2| = |2 − 3| = 1 cm
Final Answer
Correct Option: 1 (1 cm)
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