Question

Three glass slabs A, B, C of different media are arranged as shown in the figure. Refractive index of B is 1.3 and the refractive index of C is 1.7. If the number of waves passing through B and C are equal to the total number of waves passing through B and C, then the refractive index of slab A is:

• A. 1.5
• B. 1.6
• C. 1.2
• D. 1.0

Hint:

For problems involving refractive indices and wave propagation, remember that the number of waves passing through a medium is inversely proportional to its refractive index. This is crucial for solving such problems.

Answer 1

The Correct Option is A

The number of waves passing through each medium is inversely proportional to the refractive index of that medium. We are given that the number of waves passing through slabs B and C are the same.

Let the refractive indices of the slabs be \( n_A \), \( n_B \), and \( n_C \), and let the total number of waves passing through each slab be the same. Thus, we have:

\[ \frac{1}{n_A} = \frac{1}{n_B} = \frac{1}{n_C} \]

Given:
- \( n_B = 1.3 \)
- \( n_C = 1.7 \)

To find \( n_A \), we use the fact that the total number of waves passing through B and C is the same. Since the refractive index is inversely proportional to the number of waves:

\[ \frac{1}{n_A} = \frac{1}{1.3} = \frac{1}{1.7} \]

Thus, solving for \( n_A \), we find:

\[ n_A = 1.5 \]

Thus, the refractive index of slab A is \( n_A = 1.5 \).