A ray incident on the face of a prism is refracted and escape through an adjacent face. What is the maximum permissible angle of the prism, if it is made of glass with a refractive index of `mu=1.5`?

Explanation

To solve the problem of finding the maximum permissible angle of a prism made of glass with a refractive index of \( \mu = 1.5 \), we can follow these steps:

### Step 1: Understand the Concept of Critical Angle
The critical angle \( \theta_C \) is the angle of incidence above which total internal reflection occurs. It can be calculated using the formula:
\[ \theta_C = \sin^{-1}\left(\frac{1}{\mu}\right) \]
where \( \mu \) is the refractive index of the material.

### Step 2: Calculate the Critical Angle
Given \( \mu = 1.5 \):
\[ \theta_C = \sin^{-1}\left(\frac{1}{1.5}\right) = \sin^{-1}\left(\frac{2}{3}\right) \]
Using a calculator, we find:
\[ \theta_C \approx 42^\circ \]

### Step 3: Relate Angles in the Prism
For a prism, when a ray of light enters and exits, the angles of incidence and refraction can be related. Let \( r_1 \) be the angle of refraction at the first face, and \( r_2 \) be the angle of refraction at the second face. The angle \( A \) of the prism can be expressed as:
\[ A = r_1 + r_2 \]

### Step 4: Apply the Condition for Maximum Angle
To maximize the angle \( A \), we can use the symmetry of the situation. The maximum angle occurs when \( r_1 = r_2 \). Thus, we can write:
\[ A = 2r_1 \]
Since \( r_1 \) must be less than or equal to the critical angle \( \theta_C \):
\[ r_1 \leq \theta_C \]
This leads to:
\[ A \leq 2\theta_C \]

### Step 5: Substitute the Critical Angle
Substituting the value of the critical angle:
\[ A \leq 2 \times 42^\circ = 84^\circ \]

### Conclusion
The maximum permissible angle of the prism is:
\[ \boxed{84^\circ} \]