Light ray is incident on a prism of angle `A=60^(@)` are refractive index `mu=sqrt(2)`. The angle of incidence which the emergent rays grazes the surface is given

Options

A. `sin^(-1)((sqrt(3)-1)/2)`
B. `sin^(-1)((sqrt(2)-1)/2)`
C. `sin^(-1)(sqrt(3)/(2))`
D. `sin^(-1)((2)/(3sqrt(3)))`

Answer

To solve the problem, we need to find the angle of incidence \(i\) for a light ray incident on a prism with an angle \(A = 60^\circ\) and a refractive index \(\mu = \sqrt{2}\), such that the emergent ray grazes the surface of the prism.

### Step-by-Step Solution:

1. **Understanding the Emergent Ray Condition**:
Since the emergent ray grazes the surface, the angle of emergence \(E\) is \(90^\circ\).

2. **Using Snell's Law at the Emergent Surface**:
According to Snell's law at the second surface of the prism:
\[ \mu \cdot \sin(r_2) = \sin(E) \]
Here, \(E = 90^\circ\), so \(\sin(90^\circ) = 1\). Therefore:
\[ \mu \cdot \sin(r_2) = 1 \]
Substituting \(\mu = \sqrt{2}\):
\[ \sqrt{2} \cdot \sin(r_2) = 1 \]
This gives:
\[ \sin(r_2) = \frac{1}{\sqrt{2}} \implies r_2 = 45^\circ \]

3. **Relating Angles in the Prism**:
In a prism, the angles are related by:
\[ r_1 + r_2 = A \]
Substituting \(A = 60^\circ\) and \(r_2 = 45^\circ\):
\[ r_1 + 45^\circ = 60^\circ \]
Therefore:
\[ r_1 = 60^\circ - 45^\circ = 15^\circ \]

4. **Applying Snell's Law at the First Surface**:
Now, we apply Snell's law at the first surface:
\[ \sin(i) = \mu \cdot \sin(r_1) \]
Substituting \(\mu = \sqrt{2}\) and \(r_1 = 15^\circ\):
\[ \sin(i) = \sqrt{2} \cdot \sin(15^\circ) \]

5. **Calculating \(\sin(15^\circ)\)**:
We can use the sine subtraction formula:
\[ \sin(15^\circ) = \sin(45^\circ - 30^\circ) = \sin(45^\circ)\cos(30^\circ) - \cos(45^\circ)\sin(30^\circ) \]
Knowing \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\), \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), \(\cos(45^\circ) = \frac{\sqrt{2}}{2}\), and \(\sin(30^\circ) = \frac{1}{2}\):
\[ \sin(15^\circ) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6}}{4} - \frac{\sqrt{2}}{4} = \frac{\sqrt{6} - \sqrt{2}}{4} \]

6. **Substituting Back**:
Now substituting \(\sin(15^\circ)\) back into the equation for \(\sin(i)\):
\[ \sin(i) = \sqrt{2} \cdot \frac{\sqrt{6} - \sqrt{2}}{4} = \frac{\sqrt{2}(\sqrt{6} - \sqrt{2})}{4} \]

7. **Finding the Angle of Incidence**:
Finally, we find \(i\):
\[ i = \sin^{-1}\left(\frac{\sqrt{2}(\sqrt{6} - \sqrt{2})}{4}\right) \]

### Final Answer:
The angle of incidence \(i\) is given by:
\[ i = \sin^{-1}\left(\frac{\sqrt{2}(\sqrt{6} - \sqrt{2})}{4}\right) \]