Question

A glass cube of length 21 cm has a small air bubble trapped inside. When viewed normally from one face, the bubble appears to be at 12 cm. When viewed normally from the opposite face, its apparent distance is 6 cm. The refractive index of glass and the actual distance of the air bubble from the first surface respectively are \dots
Note: Based on standard physical constraints, there is a known typographical error in the standard transcript of this exam question. To yield $\mu \approx 1.5$ (glass), the apparent depth from the first face was intended to be 8 cm, not 12 cm. We will demonstrate the solution finding the closest logical answer based on standard glass refraction values.

• A. 1.5, 12 cm
• B. 1.55, 14 cm
• C. 1.6, 11 cm
• D. 1.5, 9 cm

Hint:

In competitive exams, if direct calculation yields an absurd physical value (like glass having $\mu = 1.16$), work backward from the options using the fundamental property $\text{Total Thickness} = \mu \times (\text{Sum of Apparent Depths})$. $\mu = \frac{21}{8+6} = 1.5$ restores sanity!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
An air bubble is trapped inside a glass block. Because glass is optically denser than air, the bubble appears closer to the surface than it actually is due to refraction.

Step 2: Key Formula or Approach:
The relationship between real depth, apparent depth, and the refractive index ($\mu$) is:
$$\mu = \frac{\text{Real Depth}}{\text{Apparent Depth}}$$
If the actual distances from the two opposite faces are $t_1$ and $t_2$, then $t_1 + t_2$ must equal the total length of the cube (21 cm).

Step 3: Detailed Explanation:
Let's reverse-engineer the options to find the mathematically sound scenario, given the likely typo in the question text.
Assume Option (a) is correct: $\mu = 1.5$ and the actual distance from the first surface is $t_1 = 12 \text{ cm}$.
1. Since the total block length is 21 cm, the actual distance from the opposite face is:
$$t_2 = 21 - 12 = 9 \text{ cm}.$$
2. Let's calculate what the apparent distances should be using $\mu = 1.5$:
Apparent depth from face 1: $d_1 = \frac{t_1}{\mu} = \frac{12}{1.5} = 8 \text{ cm}$.
Apparent depth from face 2: $d_2 = \frac{t_2}{\mu} = \frac{9}{1.5} = 6 \text{ cm}$.
Notice that $d_2 = 6 \text{ cm}$ perfectly matches the second condition in the question prompt!
The first condition states "appears to be at 12 cm" which is a clear typo in the question paper for "8 cm" (or they mistakenly printed the real depth $t_1$ instead of the apparent depth $d_1$ in the text).
If we forcefully used the flawed given numbers ($d_1=12, d_2=6$):
$\mu = \frac{\text{Total Real Depth}}{\text{Total Apparent Depth}} = \frac{21}{12+6} = \frac{21}{18} = 1.16$, which is the index of nothing, and fits no option.
Thus, standard testing logic dictates option (a) represents the intended true physics scenario.

Step 4: Final Answer:
The refractive index is 1.5 and actual distance is 12 cm, aligning with option (a).