Yellow light of wavelength 6000 Å 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 Å 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 step by step, we will use the formula for fringe width in Young's double slit experiment and apply the changes in wavelength and slit separation accordingly.

### Step 1: Understand the given values
- Wavelength of yellow light, \( \lambda_1 = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)
- Fringe width with yellow light, \( \beta_1 = 0.8 \, \text{mm} = 0.8 \times 10^{-3} \, \text{m} \)
- Wavelength of the new light, \( \lambda_2 = 7500 \, \text{Å} = 7500 \times 10^{-10} \, \text{m} \)
- The separation between the slits is doubled, so if the original separation is \( d_1 \), the new separation will be \( d_2 = 2d_1 \).

### Step 2: Write the formula for fringe width
The fringe width \( \beta \) in Young's double slit experiment is given by:
\[ \beta = \frac{d \lambda}{D} \]
where:
- \( d \) = separation between the slits
- \( \lambda \) = wavelength of light
- \( D \) = distance from the slits to the screen

### Step 3: Set up the equations for the two cases
For the first case (yellow light):
\[ \beta_1 = \frac{d_1 \lambda_1}{D} \]
For the second case (new light):
\[ \beta_2 = \frac{d_2 \lambda_2}{D} \]
Since \( d_2 = 2d_1 \), we can substitute this into the equation for \( \beta_2 \):
\[ \beta_2 = \frac{2d_1 \lambda_2}{D} \]

### Step 4: Relate \( \beta_2 \) to \( \beta_1 \)
We can express \( \beta_2 \) in terms of \( \beta_1 \):
\[ \frac{\beta_2}{\beta_1} = \frac{2d_1 \lambda_2 / D}{d_1 \lambda_1 / D} = \frac{2 \lambda_2}{\lambda_1} \]
Thus,
\[ \beta_2 = \beta_1 \cdot \frac{2 \lambda_2}{\lambda_1} \]

### Step 5: Substitute the known values
Now we can substitute the known values into the equation:
\[ \beta_2 = 0.8 \times 10^{-3} \cdot \frac{2 \times 7500 \times 10^{-10}}{6000 \times 10^{-10}} \]

### Step 6: Simplify the expression
Calculating the fraction:
\[ \frac{2 \times 7500}{6000} = \frac{15000}{6000} = 2.5 \]
Now substituting this back:
\[ \beta_2 = 0.8 \times 10^{-3} \cdot 2.5 = 2.0 \times 10^{-3} \, \text{m} = 2.0 \, \text{mm} \]

### Step 7: Final result
Thus, the fringe width when the source is replaced with the new wavelength and the slit separation is doubled is:
\[ \beta_2 = 2.0 \, \text{mm} \]