Question

At the interface between two materials having refractive indices \( n_1 \) and \( n_2 \), the critical angle for reflection of an EM wave is \( \theta_c \). The \( n_1 \) material is replaced by another material having refractive index \( n_3 \), such that the critical angle at the interface between \( n_1 \) and \( n_3 \) materials is \( \theta_{c3} \). If \( n_1 > n_2 > n_3 \), \( \frac{n_2}{n_3} = \frac{2}{5} \), and \( \sin \theta_{c2} - \sin \theta_{c1} = \frac{1}{2} \), then \( \theta_{c1} \) is:

• A. $\sin^{-1} \left( -\frac{5}{6} \right)$
• B. \( \sin^{-1} \left( \frac{2}{3n_1} \right) \)
• C. \( \sin^{-1} \left( \frac{1}{3n_1} \right) \)
• D. \( \sin^{-1} \left( \frac{1}{6n_1} \right) \)

Hint:

For problems involving critical angles and refractive indices, use Snell's Law and the relationship between the angles to find the unknowns.

Answer 1

The Correct Option is A

We are tasked with solving for the critical angles $ \theta_{1C} $ and $ \theta_{2C} $, and determining the relationship between the refractive indices $ n_1 $, $ n_2 $, and $ n_3 $. The solution proceeds as follows:

1. Critical Angle Relations:
The sine of the critical angle is given by:

$ \sin \theta_{1C} = \frac{n_1}{n_2} $

$ \sin \theta_{2C} = \frac{n_1}{n_3} $

2. Difference Between Sines:
We are given that:

$ \sin \theta_{2C} - \sin \theta_{1C} = \frac{1}{2} $

Substituting the expressions for $ \sin \theta_{1C} $ and $ \sin \theta_{2C} $:

$ \frac{n_1}{n_3} - \frac{n_1}{n_2} = \frac{1}{2} $

3. Simplifying the Equation:
Factor out $ n_1 $:

$ n_1 \left( \frac{1}{n_3} - \frac{1}{n_2} \right) = \frac{1}{2} $

Simplify the terms inside the parentheses:

$ \frac{1}{n_3} - \frac{1}{n_2} = \frac{n_2 - n_3}{n_2 n_3} $

Thus, the equation becomes:

$ n_1 \cdot \frac{n_2 - n_3}{n_2 n_3} = \frac{1}{2} $

4. Substituting Given Values:
From the problem, we know:

$ n_1 \left( \frac{2}{5} - 1 \right) = \frac{n_3}{2} $

Simplify $ \frac{2}{5} - 1 $:

$ \frac{2}{5} - 1 = \frac{2}{5} - \frac{5}{5} = -\frac{3}{5} $

Substitute this back into the equation:

$ n_1 \cdot \left( -\frac{3}{5} \right) = \frac{n_3}{2} $

Solve for $ n_1 $:

$ n_1 = \frac{n_3}{2} \cdot \left( -\frac{5}{3} \right) = -\frac{5}{6} n_3 $

5. Ratio of Refractive Indices:
We now determine the ratio $ \frac{n_1}{n_2} $. From earlier, we know:

$ \frac{n_1}{n_2} = \frac{-5}{6} $

6. Inverse Sine Calculation:
Finally, the critical angle $ \theta_{1C} $ is given by:

$ \theta_{1C} = \sin^{-1} \left( \frac{n_1}{n_2} \right) $

Substitute $ \frac{n_1}{n_2} = \frac{-5}{6} $:

$ \theta_{1C} = \sin^{-1} \left( -\frac{5}{6} \right) $

Final Answer:
The critical angle $ \theta_{1C} $ is: $ \sin^{-1} \left( -\frac{5}{6} \right) $