In Young's experiment for interference of light the slits 0.2 cm apart are illuminated by yellow light `(lamda=5896A^(@))` What would be the fringe width on a screen placed 1 m from the plane of slits? What will be the fringe width if the system is immersed in water (refractive index `=4//3`)

Explanation

To solve the problem step by step, we will calculate the fringe width in Young's double-slit experiment both in air and when the system is immersed in water.

### Given Data:
- Wavelength of yellow light, \( \lambda = 5896 \, \text{Å} = 5896 \times 10^{-10} \, \text{m} \)
- Distance between the slits, \( d = 0.2 \, \text{cm} = 0.2 \times 10^{-2} \, \text{m} = 2 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen, \( D = 1 \, \text{m} \)
- Refractive index of water, \( n = \frac{4}{3} \)

### Step 1: Calculate the fringe width in air
The formula for fringe width \( \beta \) in Young's experiment is given by:
\[ \beta = \frac{\lambda D}{d} \]

Substituting the values:
\[ \beta = \frac{5896 \times 10^{-10} \, \text{m} \times 1 \, \text{m}}{2 \times 10^{-3} \, \text{m}} \]

Calculating this gives:
\[ \beta = \frac{5896 \times 10^{-10}}{2 \times 10^{-3}} = 2948 \times 10^{-7} \, \text{m} = 2.948 \times 10^{-4} \, \text{m} = 0.0002948 \, \text{m} = 0.2948 \, \text{mm} \]

### Step 2: Calculate the fringe width in water
When the system is immersed in water, the wavelength of light changes according to the refractive index:
\[ \lambda' = \frac{\lambda}{n} \]

Substituting the values:
\[ \lambda' = \frac{5896 \times 10^{-10} \, \text{m}}{\frac{4}{3}} = 5896 \times 10^{-10} \times \frac{3}{4} = 4422 \times 10^{-10} \, \text{m} \]

Now, we can calculate the new fringe width \( \beta' \) in water using the same formula:
\[ \beta' = \frac{\lambda' D}{d} \]

Substituting the values:
\[ \beta' = \frac{4422 \times 10^{-10} \, \text{m} \times 1 \, \text{m}}{2 \times 10^{-3} \, \text{m}} \]

Calculating this gives:
\[ \beta' = \frac{4422 \times 10^{-10}}{2 \times 10^{-3}} = 2211 \times 10^{-7} \, \text{m} = 2.211 \times 10^{-4} \, \text{m} = 0.0002211 \, \text{m} = 0.2211 \, \text{mm} \]

### Final Answers:
- Fringe width in air: \( \beta = 0.2948 \, \text{mm} \)
- Fringe width in water: \( \beta' = 0.2211 \, \text{mm} \)