A light ray is incident on a transparent sphere of index = `sqrt(2)` , at an angle of incidence =`45^(@)` , What is the deviation of a tiny fraction of the ray, which enters the sphere, undergoes two internal reflections and then refracts out into air?

Options

A. `270^(@)`
B. `240^(@)`
C. `120^(@)`
D. `180^(@)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Determine the angle of refraction at the first interface
Given:
- Refractive index of the sphere, \( \mu = \sqrt{2} \)
- Angle of incidence, \( i = 45^\circ \)
- Refractive index of air, \( \mu_1 = 1 \)

Using Snell's law:
\[ \mu_1 \sin(i) = \mu_2 \sin(r) \]
Substituting the values:
\[ 1 \cdot \sin(45^\circ) = \sqrt{2} \cdot \sin(r) \]
\[ \sin(r) = \frac{\sin(45^\circ)}{\sqrt{2}} = \frac{\frac{\sqrt{2}}{2}}{\sqrt{2}} = \frac{1}{2} \]
Thus,
\[ r = 30^\circ \]

### Step 2: Analyze the internal reflections
After refraction, the ray enters the sphere and will undergo two internal reflections. The angle of incidence for each reflection will be equal to the angle of refraction from the previous interface.

1. **First internal reflection**: The angle of incidence is \( 30^\circ \), so the angle of reflection is also \( 30^\circ \).
2. **Second internal reflection**: The angle of incidence remains \( 30^\circ \), leading to another reflection at \( 30^\circ \).

### Step 3: Determine the angle of refraction at the exit interface
After the second reflection, the ray exits the sphere. The angle of incidence at the exit interface is again \( 30^\circ \).

Using Snell's law again:
\[ \sqrt{2} \cdot \sin(30^\circ) = 1 \cdot \sin(r') \]
\[ \sin(r') = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2} \]
Thus,
\[ r' = 45^\circ \]

### Step 4: Calculate the total deviation
Now, we need to calculate the total deviation caused by the two refractions and two reflections.

1. **Deviation due to refraction**: Each refraction causes a deviation of \( 15^\circ \) (since \( 45^\circ - 30^\circ = 15^\circ \)). Thus, for two refractions:
\[ \text{Total deviation due to refraction} = 2 \times 15^\circ = 30^\circ \]

2. **Deviation due to reflection**: Each reflection also contributes to the deviation. The deviation caused by each reflection is \( 30^\circ \) (the angle of incidence). Thus, for two reflections:
\[ \text{Total deviation due to reflection} = 2 \times 30^\circ = 60^\circ \]

### Step 5: Combine the deviations
The net deviation \( D \) is given by:
\[ D = \text{Deviation due to refraction} + \text{Deviation due to reflection} \]
\[ D = 30^\circ + 60^\circ = 90^\circ \]

### Final Result
The total deviation of the light ray after undergoing two internal reflections and then refracting out into air is \( 90^\circ \).

---