The angles of refraction of a very thin prism is `1^(@)`. A light say is incident normally on one of the refracting surface. The ray that ultimately emerges from the first surface, after suffering from the second surface. Makes an angle of `3.32^(@)` with the normal. the deviation of the ray emerging from the second surface and the refractive index of the material of the prism respectivley, are

Options

A. a.`0.66^(@),1.66`
B. b.`1.66^(@),1.5`
C. c.`1.5^(@),1.66`
D. d.`0.66^(@),1.5`

Explanation

To solve the problem, we need to find the deviation of the ray emerging from the second surface of a thin prism and the refractive index of the material of the prism.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The angle of refraction (r) at the first surface of the prism is given as \(1^\circ\).
- The angle made by the emerging ray with the normal after passing through the second surface is \(3.32^\circ\).
- We need to find the deviation (D) and the refractive index (n) of the prism.

2. **Determine the Angles**:
- When light enters the prism normally, the angle of incidence (i) is \(0^\circ\), and thus the angle of refraction at the first surface is \(1^\circ\).
- The angle of deviation (D) can be calculated using the formula:
\[ D = i + r_2 - A \]
where \(A\) is the angle of the prism, which we can find using the geometry of the prism.

3. **Calculate the Angle of the Prism (A)**:
- For a very thin prism, the angle of the prism \(A\) can be approximated as:
\[ A = r + r_2 = 1^\circ + 3.32^\circ = 4.32^\circ \]

4. **Calculate the Deviation (D)**:
- The deviation can be calculated using:
\[ D = i + r_2 - A \]
- Substituting the values:
\[ D = 0^\circ + 3.32^\circ - 4.32^\circ = -1^\circ \]
- Since deviation is typically taken as a positive quantity, we take the absolute value:
\[ D = 1^\circ \]

5. **Calculate the Refractive Index (n)**:
- Using Snell's law at the first surface:
\[ n \cdot \sin(i) = \sin(r) \]
- Since \(i = 0^\circ\), we can write:
\[ n \cdot \sin(1^\circ) = \sin(3.32^\circ) \]
- For small angles, \(\sin(\theta) \approx \theta\) in radians:
\[ n \cdot 1^\circ \approx 3.32^\circ \]
- Converting degrees to radians:
\[ n \cdot \frac{\pi}{180} \approx \frac{3.32 \cdot \pi}{180} \]
- Thus, we can find \(n\):
\[ n = \frac{3.32}{1} \cdot \frac{180}{\pi} \approx 1.66 \]

### Final Results:
- The deviation \(D\) is \(1^\circ\).
- The refractive index \(n\) is approximately \(1.66\).