In Young's experiment interference bands are produced on the screen placed at `1.5m` from the slits `0.15mm` apart and illuminated by light of wavelength `6000 Å`. If the screen is now taken away from the slit by 50 cm the change in the fringe width will be

Answer

To solve the problem, we need to calculate the change in fringe width when the distance of the screen from the slits is altered. We will follow these steps:

### Step 1: Understand the formula for fringe width
The fringe width (β) in Young's double-slit experiment is given by the formula:
\[ \beta = \frac{\lambda D}{d} \]
where:
- \( \lambda \) = wavelength of light
- \( D \) = distance from the slits to the screen
- \( d \) = distance between the slits

### Step 2: Identify the initial parameters
From the problem, we have:
- Initial distance from the slits to the screen, \( D_1 = 1.5 \, m \)
- Distance between the slits, \( d = 0.15 \, mm = 0.15 \times 10^{-3} \, m \)
- Wavelength of light, \( \lambda = 6000 \, Å = 6000 \times 10^{-10} \, m = 6 \times 10^{-7} \, m \)

### Step 3: Calculate the initial fringe width (β1)
Using the formula:
\[ \beta_1 = \frac{\lambda D_1}{d} \]
Substituting the known values:
\[ \beta_1 = \frac{(6 \times 10^{-7}) \times (1.5)}{0.15 \times 10^{-3}} \]

### Step 4: Calculate the new distance
The screen is moved away by 50 cm, so the new distance from the slits to the screen is:
\[ D_2 = D_1 + 0.5 \, m = 1.5 + 0.5 = 2.0 \, m \]

### Step 5: Calculate the new fringe width (β2)
Using the new distance:
\[ \beta_2 = \frac{\lambda D_2}{d} \]
Substituting the new values:
\[ \beta_2 = \frac{(6 \times 10^{-7}) \times (2.0)}{0.15 \times 10^{-3}} \]

### Step 6: Calculate the change in fringe width (Δβ)
The change in fringe width is given by:
\[ \Delta \beta = \beta_2 - \beta_1 \]
Substituting the expressions for β1 and β2:
\[ \Delta \beta = \frac{\lambda D_2}{d} - \frac{\lambda D_1}{d} \]
\[ \Delta \beta = \frac{\lambda (D_2 - D_1)}{d} \]
Substituting the values:
\[ \Delta \beta = \frac{(6 \times 10^{-7}) \times (2.0 - 1.5)}{0.15 \times 10^{-3}} \]
\[ \Delta \beta = \frac{(6 \times 10^{-7}) \times (0.5)}{0.15 \times 10^{-3}} \]

### Step 7: Simplify the expression
Calculating the above expression:
\[ \Delta \beta = \frac{3 \times 10^{-7}}{0.15 \times 10^{-3}} = \frac{3 \times 10^{-7}}{1.5 \times 10^{-4}} = 2 \times 10^{-3} \, m \]

### Final Answer
The change in fringe width is:
\[ \Delta \beta = 2 \times 10^{-3} \, m \]

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