Yellow light of wavelength 6000 Ã… produces fringes of width 0.8 mm in Young's double slit experiment. If the source is replaced by another monochromatic source of wavelength 7500 Ã… and the separation between the slits is doubled then the fringe width becomes

Options

A. 0.1 mm
B. 0.5 mm
C. 4.3 mm
D. 1 mm

Explanation

To solve the problem, we will follow these steps systematically:

### 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,
- \( d \) = separation between the slits.

### Step 2: Calculate the initial fringe width (β₁)
In the first case, we have:
- Wavelength \( \lambda_1 = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)
- Fringe width \( \beta_1 = 0.8 \, \text{mm} = 0.8 \times 10^{-3} \, \text{m} \)

Using the formula for fringe width:
\[ \beta_1 = \frac{\lambda_1 D}{d} \]

### Step 3: Set up the equation for the second case (β₂)
In the second case, we have:
- Wavelength \( \lambda_2 = 7500 \, \text{Å} = 7500 \times 10^{-10} \, \text{m} \)
- The separation between the slits is doubled, so the new separation \( d' = 2d \).

The new fringe width \( \beta_2 \) can be expressed as:
\[ \beta_2 = \frac{\lambda_2 D}{d'} \]
Substituting \( d' = 2d \):
\[ \beta_2 = \frac{\lambda_2 D}{2d} \]

### Step 4: Relate β₂ to β₁
Now, we can relate \( \beta_2 \) to \( \beta_1 \):
\[ \beta_2 = \frac{\lambda_2 D}{2d} = \frac{\lambda_2}{2} \cdot \frac{D}{d} = \frac{\lambda_2}{2} \cdot \frac{d}{\lambda_1} \cdot \beta_1 \]
Thus,
\[ \frac{\beta_2}{\beta_1} = \frac{\lambda_2}{2 \lambda_1} \]

### Step 5: Substitute the known values
Now substituting the known values:
\[ \frac{\beta_2}{0.8 \times 10^{-3}} = \frac{7500 \times 10^{-10}}{2 \times 6000 \times 10^{-10}} \]
This simplifies to:
\[ \frac{\beta_2}{0.8 \times 10^{-3}} = \frac{7500}{12000} \]
Calculating the fraction:
\[ \frac{7500}{12000} = \frac{5}{8} \]

### Step 6: Calculate β₂
Now, we can find \( \beta_2 \):
\[ \beta_2 = 0.8 \times 10^{-3} \times \frac{5}{8} \]
Calculating this:
\[ \beta_2 = 0.8 \times \frac{5}{8} \times 10^{-3} = 0.5 \times 10^{-3} \, \text{m} = 0.5 \, \text{mm} \]

### Final Answer
Thus, the new fringe width \( \beta_2 \) becomes:
\[ \beta_2 = 0.5 \, \text{mm} \]