The maximum value of refractive index of a prism which permits the transmission of light through it when the refracting angl e of the prism is `90^@,` is given by

Options

A. `1.500`
B. `1.414`
C. `2.000`
D. `1.732`

Explanation

To find the maximum value of the refractive index (μ) of a prism that permits the transmission of light through it when the refracting angle (A) of the prism is 90 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Geometry of the Prism**:
- The refracting angle of the prism is given as \( A = 90^\circ \).
- The prism has two refracting surfaces, and we denote the angles of refraction at these surfaces as \( R_1 \) and \( R_2 \).
- According to the geometry of the prism, we have:
\[ R_1 + R_2 = A \]
- Since \( A = 90^\circ \), we can write:
\[ R_1 + R_2 = 90^\circ \]

2. **Using Symmetry for Maximum Refractive Index**:
- For maximum refractive index, we assume symmetry, which means \( R_1 = R_2 \).
- Let \( R_1 = R_2 = R \).
- Therefore, we can rewrite the equation as:
\[ 2R = 90^\circ \]
- Solving for \( R \):
\[ R = 45^\circ \]

3. **Condition for Light Transmission**:
- For light to pass through the prism without total internal reflection, the angle of refraction \( R \) must not exceed the critical angle \( \theta_c \).
- Thus, we have:
\[ R \leq \theta_c \]
- Substituting \( R = 45^\circ \):
\[ 45^\circ \leq \theta_c \]

4. **Relating Critical Angle to Refractive Index**:
- The critical angle \( \theta_c \) is related to the refractive index \( \mu \) by the formula:
\[ \theta_c = \sin^{-1}\left(\frac{1}{\mu}\right) \]
- Therefore, we can express the condition as:
\[ 45^\circ \leq \sin^{-1}\left(\frac{1}{\mu}\right) \]

5. **Finding the Sine of the Critical Angle**:
- Taking the sine of both sides:
\[ \sin(45^\circ) \leq \frac{1}{\mu} \]
- We know that \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \), so we can rewrite the inequality as:
\[ \frac{1}{\sqrt{2}} \leq \frac{1}{\mu} \]

6. **Rearranging to Find Maximum Refractive Index**:
- Rearranging gives us:
\[ \mu \leq \sqrt{2} \]
- Therefore, the maximum value of the refractive index \( \mu \) is:
\[ \mu = \sqrt{2} \]

### Conclusion:
The maximum value of the refractive index of the prism that permits the transmission of light through it when the refracting angle is \( 90^\circ \) is \( \sqrt{2} \).