The refractive index of a prism is `mu`. Find the maximum angle of the prism for which a ray incident on it will be transmitted through other face without total internal reflection.

Explanation

To find the maximum angle of a prism for which a ray incident on it will be transmitted through the other face without total internal reflection (TIR), we can follow these steps:

### Step 1: Understand the Conditions for Total Internal Reflection
Total internal reflection occurs when the angle of incidence in a denser medium exceeds the critical angle. For a ray to pass through the prism without TIR, the angle of refraction at the second face of the prism must be less than the critical angle.

### Step 2: Define the Angles
Let:
- \( A \) = angle of the prism
- \( I \) = angle of incidence at the first face
- \( R \) = angle of refraction at the first face
- \( R' \) = angle of incidence at the second face
- \( \mu \) = refractive index of the prism

### Step 3: Apply Snell's Law
At the first face of the prism, we apply Snell's law:
\[ n_1 \sin I = n_2 \sin R \]
Assuming the light is coming from air (where \( n_1 = 1 \)) into the prism (where \( n_2 = \mu \)):
\[ \sin I = \mu \sin R \tag{1} \]

### Step 4: Relate Angles of Refraction and Incidence
From the geometry of the prism, we know:
\[ R + R' = A \tag{2} \]
Where \( R' \) is the angle of incidence at the second face.

### Step 5: Condition for No Total Internal Reflection
For no total internal reflection at the second face, the angle of incidence \( R' \) must be less than the critical angle \( I_C \). The critical angle can be defined as:
\[ I_C = \sin^{-1} \left( \frac{1}{\mu} \right) \]
Thus, we have:
\[ R' < I_C \tag{3} \]

### Step 6: Substitute and Rearrange
From equation (2), we can express \( R' \) in terms of \( R \):
\[ R' = A - R \]
Substituting this into inequality (3):
\[ A - R < \sin^{-1} \left( \frac{1}{\mu} \right) \]
Rearranging gives:
\[ A < R + \sin^{-1} \left( \frac{1}{\mu} \right) \tag{4} \]

### Step 7: Use Snell's Law Again
From equation (1), we can express \( R \) in terms of \( I \):
\[ R = \sin^{-1} \left( \frac{\sin I}{\mu} \right) \]
To find the maximum angle \( A \), we need to maximize \( R \). The maximum occurs when \( I \) approaches \( 90^\circ \):
\[ R \to \sin^{-1} \left( \frac{1}{\mu} \right) \]

### Step 8: Final Expression for Maximum Angle
Substituting this back into equation (4):
\[ A < \sin^{-1} \left( \frac{1}{\mu} \right) + \sin^{-1} \left( \frac{1}{\mu} \right) \]
Thus, the maximum angle \( A_{max} \) is:
\[ A_{max} = 2 \sin^{-1} \left( \frac{1}{\mu} \right) \]

### Conclusion
The maximum angle of the prism for which a ray incident on it will be transmitted through the other face without total internal reflection is:
\[ A_{max} = 2 \sin^{-1} \left( \frac{1}{\mu} \right) \]

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