Light falls at normal incidence on one face of a glass prism of refractive index `sqrt(2)`. Then the angle of emergence when the angle of the prism is `45^@`

Options

A. 45°
B. 60°
C. 15°
D. 90°

Explanation

To solve the problem, we need to find the angle of emergence when light falls normally on one face of a glass prism with a refractive index of \(\sqrt{2}\) and an angle of the prism of \(45^\circ\).

### Step-by-step Solution:

1. **Understanding the Setup**:
- The light ray falls normally on one face of the prism. This means that the angle of incidence \(i = 0^\circ\).
- The angle of the prism \(A = 45^\circ\).

2. **Refraction at the First Face**:
- Since the light is incident normally, it will pass through the first face without bending. Therefore, the angle of refraction \(R_1 = 0^\circ\).

3. **Using the Prism Formula**:
- For a prism, the relationship between the angle of the prism \(A\), the angle of refraction \(R_1\), and the angle of refraction \(R_2\) at the second face is given by:
\[ A = R_1 + R_2 \]
- Substituting the known values:
\[ 45^\circ = 0^\circ + R_2 \implies R_2 = 45^\circ \]

4. **Applying Snell's Law at the Second Face**:
- We apply Snell's law at the second face of the prism:
\[ n_1 \sin R_2 = n_2 \sin E \]
- Here, \(n_1 = \sqrt{2}\) (the refractive index of the prism), \(n_2 = 1\) (the refractive index of air), and \(R_2 = 45^\circ\).
- Substituting the values:
\[ \sqrt{2} \sin(45^\circ) = 1 \sin E \]
- Since \(\sin(45^\circ) = \frac{1}{\sqrt{2}}\):
\[ \sqrt{2} \cdot \frac{1}{\sqrt{2}} = \sin E \implies 1 = \sin E \]

5. **Finding the Angle of Emergence**:
- The angle \(E\) such that \(\sin E = 1\) is:
\[ E = 90^\circ \]

### Final Answer:
The angle of emergence \(E\) is \(90^\circ\).

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