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. `0.15^(@)`
B. `1^(@)`
C. `2^(@)`
D. `0.3^(@)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between angular width and wavelength
In a double slit experiment, the angular width \( \theta \) of the fringe is given by the formula:
\[ \theta = \frac{\lambda}{d} \]
where \( \lambda \) is the wavelength of light and \( d \) is the distance between the slits.

### Step 2: Calculate the initial angular width
Given that the initial angular width \( \theta \) is \( 0.2^\circ \) and the wavelength \( \lambda \) is \( 600 \, \text{nm} \) (or \( 600 \times 10^{-9} \, \text{m} \)), we can express \( d \) in terms of \( \theta \):
\[ d = \frac{\lambda}{\theta} \]
To use this formula, we need to convert \( \theta \) from degrees to radians:
\[ \theta = 0.2^\circ \times \frac{\pi}{180} \approx 0.00349 \, \text{radians} \]
Now, substituting the values:
\[ d = \frac{600 \times 10^{-9}}{0.00349} \approx 1.72 \times 10^{-4} \, \text{m} \]

### Step 3: Determine the new wavelength in water
When the apparatus is immersed in water, the wavelength of light changes. The new wavelength \( \lambda' \) in water is given by:
\[ \lambda' = \frac{\lambda}{n} \]
where \( n \) is the refractive index of water (given as \( \frac{4}{3} \)):
\[ \lambda' = \frac{600 \times 10^{-9}}{\frac{4}{3}} = 600 \times 10^{-9} \times \frac{3}{4} = 450 \times 10^{-9} \, \text{m} \]

### Step 4: Calculate the new angular width in water
Using the same formula for angular width in water:
\[ \theta' = \frac{\lambda'}{d} \]
Substituting \( \lambda' \) and \( d \):
\[ \theta' = \frac{450 \times 10^{-9}}{1.72 \times 10^{-4}} \approx 0.00261 \, \text{radians} \]
Convert this back to degrees:
\[ \theta' = 0.00261 \times \frac{180}{\pi} \approx 0.149 \, \text{degrees} \]

### Step 5: Final result
The new angular width of the fringe when the apparatus is immersed in water is approximately:
\[ \theta' \approx 0.15^\circ \]

### Summary
The angular width of the fringe when the entire experimental apparatus is immersed in water is approximately \( 0.15^\circ \).

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