A thin glass prism has a refracting angle of `6^(@)` . The angle of incidence is very small . What is the deviation produced by the prism , if the prism is kept in water ? `[""_(a) n_(g) = 1.5 , "" _(a) n_(w) = 1.33`]

Options

A. `0.6^(@)`
B. `0.192^(@)`
C. `0.75^(@)`
D. `0.8^(@)`

Explanation

To solve the problem of finding the deviation produced by a thin glass prism kept in water, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Given Data:**
- Refracting angle of the prism, \( A = 6^\circ \)
- Refractive index of glass, \( n_g = 1.5 \)
- Refractive index of water, \( n_w = 1.33 \)

2. **Calculate the Refractive Index of Glass with respect to Water:**
- The formula for the refractive index of glass with respect to water is given by:
\[ \mu_{g/w} = \frac{n_g}{n_w} \]
- Substituting the values:
\[ \mu_{g/w} = \frac{1.5}{1.33} \approx 1.127 \]

3. **Use the Deviation Formula:**
- The formula for the deviation \( \delta \) produced by the prism when the angle of incidence is very small is:
\[ \delta = A \left( \mu_{g/w} - 1 \right) \]
- Substitute \( A = 6^\circ \) and \( \mu_{g/w} \approx 1.127 \):
\[ \delta = 6 \left( 1.127 - 1 \right) \]
- Calculate:
\[ \delta = 6 \times 0.127 \approx 0.762^\circ \]

4. **Final Result:**
- The deviation produced by the prism when kept in water is approximately \( 0.762^\circ \).