Question:
A rectangular glass block of thickness 10 cm and refractive index 1.5 is placed over a small coin. A beaker is filled with water of refractive index 4/3 to a height of 10 cm and placed over the glass block. The apparent depth of coin when viewed at near normal incidence is:
A rectangular glass block of thickness 10 cm and refractive index 1.5 is placed over a small coin. A beaker is filled with water of refractive index 4/3 to a height of 10 cm and placed over the glass block. The apparent depth of coin when viewed at near normal incidence is:
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Apparent depth is additive; calculate the apparent depth for each layer independently and sum them up.
Updated On: Jun 9, 2026
- \( 3.3 \text{ cm} \)
- \( 5.8 \text{ cm} \)
- \( 12.0 \text{ cm} \)
- \( 14.2 \text{ cm} \)
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The Correct Option is D
Solution and Explanation
Concept:
When looking through multiple media, the apparent depth is the sum of the apparent shifts (or apparent depths) caused by each individual medium. The apparent depth of a slab is given by \( d_{apparent} = \frac{d_{real}}{\mu} \).
Step 1: Calculate apparent depth through the glass block.
Real thickness of glass \( t_g = 10 \text{ cm} \), \( \mu_g = 1.5 \). $$ d_1 = \frac{10}{1.5} = \frac{10}{3/2} = 6.67 \text{ cm} $$
Step 2: Calculate apparent depth through the water.
Real thickness of water \( t_w = 10 \text{ cm} \), \( \mu_w = 4/3 \). $$ d_2 = \frac{10}{4/3} = \frac{30}{4} = 7.5 \text{ cm} $$
Step 3: Total apparent depth.
$$ d_{total} = d_1 + d_2 = 6.67 + 7.5 = 14.17 \text{ cm} \approx 14.2 \text{ cm} $$ $$\boxed{14.2 \text{ cm}}$$
Step 1: Calculate apparent depth through the glass block.
Real thickness of glass \( t_g = 10 \text{ cm} \), \( \mu_g = 1.5 \). $$ d_1 = \frac{10}{1.5} = \frac{10}{3/2} = 6.67 \text{ cm} $$
Step 2: Calculate apparent depth through the water.
Real thickness of water \( t_w = 10 \text{ cm} \), \( \mu_w = 4/3 \). $$ d_2 = \frac{10}{4/3} = \frac{30}{4} = 7.5 \text{ cm} $$
Step 3: Total apparent depth.
$$ d_{total} = d_1 + d_2 = 6.67 + 7.5 = 14.17 \text{ cm} \approx 14.2 \text{ cm} $$ $$\boxed{14.2 \text{ cm}}$$
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