A ray of light passes from a medium A having refractive index 1.6 to the medium B having refractive index 1.5. The value of critical angle of medium
A is….

Options

A. `sin^(-1) sqrt((16)/(15))`
B. `sin^(-1) ((16)/(15))`
C. `sin^(-1) ((1)/(2))`
D. `sin^(-1) ((15)/(16))`

Explanation

To find the critical angle for light passing from medium A (with a refractive index of 1.6) to medium B (with a refractive index of 1.5), we can use Snell's law, which states:

\[ n_1 \sin \theta_1 = n_2 \sin \theta_2 \]

Where:
- \( n_1 \) is the refractive index of medium A (1.6),
- \( n_2 \) is the refractive index of medium B (1.5),
- \( \theta_1 \) is the angle of incidence (which will be the critical angle \( \theta_c \)),
- \( \theta_2 \) is the angle of refraction (which will be 90 degrees at the critical angle).

### Step-by-Step Solution:

1. **Identify the refractive indices**:
- Refractive index of medium A, \( n_1 = 1.6 \)
- Refractive index of medium B, \( n_2 = 1.5 \)

2. **Set up Snell's law for the critical angle**:
- At the critical angle, the angle of refraction \( \theta_2 \) is 90 degrees. Therefore, we can write:
\[ n_1 \sin \theta_c = n_2 \sin 90^\circ \]

3. **Substitute the values into the equation**:
- Since \( \sin 90^\circ = 1 \), we have:
\[ 1.6 \sin \theta_c = 1.5 \times 1 \]

4. **Rearrange the equation to solve for \( \sin \theta_c \)**:
\[ \sin \theta_c = \frac{1.5}{1.6} \]

5. **Calculate \( \sin \theta_c \)**:
- Simplifying \( \frac{1.5}{1.6} \):
\[ \sin \theta_c = \frac{15}{16} \]

6. **Find the critical angle \( \theta_c \)**:
- To find \( \theta_c \), take the inverse sine:
\[ \theta_c = \sin^{-1}\left(\frac{15}{16}\right) \]

### Final Answer:
The critical angle \( \theta_c \) for medium A is \( \sin^{-1}\left(\frac{15}{16}\right) \).

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