In a double slit experiment the angular width of a fringe is found to be `0.2^(@)` on a screen placed I m away. The wavelength of light used in 600 nm. What will be the angular width of the fringe if the entire experimental apparatus is immersed in water ? Take refractive index of water to be `4//3`.

Options

A. `0.2^(@)`
B. `0.27^(@)`
C. `0.3^(@)`
D. `0.15^(@)`

Answer

To solve the problem, we need to find the angular width of the fringe when the entire experimental apparatus is immersed in water. We will use the relationship between the wavelength of light in air and in water, as well as the angular width of the fringes.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Wavelength of light in air, \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Angular width of the fringe in air, \( \theta = 0.2^\circ \)
- Refractive index of water, \( \mu = \frac{4}{3} \)
- Distance from the slits to the screen, \( D = 1 \, \text{m} \)

2. **Convert Angular Width to Radians:**
- Since calculations are often easier in radians, convert \( \theta \) from degrees to radians:
\[ \theta = 0.2^\circ \times \frac{\pi}{180} \approx 0.00349 \, \text{radians} \]

3. **Calculate Wavelength in Water:**
- The wavelength of light in a medium is given by:
\[ \lambda' = \frac{\lambda}{\mu} \]
- Substituting the values:
\[ \lambda' = \frac{600 \times 10^{-9}}{\frac{4}{3}} = 600 \times 10^{-9} \times \frac{3}{4} = 450 \times 10^{-9} \, \text{m} \]

4. **Relate Angular Widths:**
- The angular width of the fringe in air and in water can be related by:
\[ \frac{\lambda}{\theta} = \frac{\lambda'}{\theta'} \]
- Rearranging gives:
\[ \theta' = \theta \times \frac{\lambda'}{\lambda} \]

5. **Substituting Values:**
- Now substitute \( \lambda' \) and \( \lambda \):
\[ \theta' = \theta \times \frac{450 \times 10^{-9}}{600 \times 10^{-9}} = \theta \times \frac{3}{4} \]
- Substitute \( \theta = 0.2^\circ \):
\[ \theta' = 0.2^\circ \times \frac{3}{4} = 0.15^\circ \]

6. **Final Answer:**
- The angular width of the fringe when the apparatus is immersed in water is:
\[ \theta' = 0.15^\circ \]