The maximum refractive index of a material, of a prism of apex `90^(@)` , for which light may be transmitted is

Options

A. `sqrt3`
B. `1.5`
C. `sqrt2`
D. none of these

Explanation

To find the maximum refractive index of a material of a prism with an apex angle of \(90^\circ\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Prism Configuration**:
- We have a prism with an apex angle \(A = 90^\circ\).
- When light enters the prism, it refracts at the first surface and then refracts again at the second surface.

2. **Using the Refractive Angles**:
- Let \(r_1\) be the angle of refraction at the first surface and \(r_2\) be the angle of refraction at the second surface.
- According to the geometry of the prism, we have:
\[ r_1 + r_2 = A = 90^\circ \]

3. **Critical Angle Condition**:
- For light to be transmitted through the prism, the angles of refraction must not exceed the critical angle \( \theta_c \) for the material of the prism.
- Since \(r_1 = r_2 = \theta_c\) at maximum refractive index, we can set:
\[ r_1 = r_2 = \theta_c \]

4. **Setting Up the Equation**:
- From the equation \(r_1 + r_2 = 90^\circ\), substituting \(r_1\) and \(r_2\) gives:
\[ \theta_c + \theta_c = 90^\circ \implies 2\theta_c = 90^\circ \implies \theta_c = 45^\circ \]

5. **Using the Critical Angle Formula**:
- The critical angle is related to the refractive index \(n\) of the material by the formula:
\[ \sin(\theta_c) = \frac{1}{n} \]
- Substituting \(\theta_c = 45^\circ\):
\[ \sin(45^\circ) = \frac{1}{n} \]
- Since \(\sin(45^\circ) = \frac{1}{\sqrt{2}}\):
\[ \frac{1}{\sqrt{2}} = \frac{1}{n} \]

6. **Solving for the Refractive Index**:
- Rearranging the equation gives:
\[ n = \sqrt{2} \]

### Final Answer:
The maximum refractive index of the material of the prism is \(n = \sqrt{2}\).

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