In Young's double slit experiment, the fringe width is ` 1 xx 10^(-4) m` if the distance between the slit and screen is doubled and the distance between the two slit is reduced to half and wavelength is changed from `6.4 xx 10^(7) m` to `4.0 xx 10^(-7)m` , the value of new fringe width will be

Options

A. `2.5xx10^(-4)m`
B. `2.0xx10^(-4)m`
C. `1xx10^(-4)m`
D. `0.5xx10^(-4)m`

Explanation

To solve the problem, we will use the formula for fringe width in Young's double slit experiment, which is given by:

\[ \beta = \frac{\lambda D}{d} \]

where:
- \(\beta\) is the fringe width,
- \(\lambda\) is the wavelength of light,
- \(D\) is the distance between the slits and the screen,
- \(d\) is the distance between the two slits.

### Step 1: Identify the initial conditions
We are given:
- Initial fringe width, \(\beta_1 = 1 \times 10^{-4} \, \text{m}\)
- Initial wavelength, \(\lambda_1 = 6.4 \times 10^{-7} \, \text{m}\)
- Initial distance between the slits and the screen, \(D_1\)
- Initial distance between the slits, \(d_1\)

### Step 2: Determine the new conditions
According to the problem:
- The distance between the slits and the screen is doubled: \(D_2 = 2D_1\)
- The distance between the two slits is halved: \(d_2 = \frac{d_1}{2}\)
- The new wavelength is: \(\lambda_2 = 4.0 \times 10^{-7} \, \text{m}\)

### Step 3: Write the expression for new fringe width
Using the formula for fringe width, the new fringe width \(\beta_2\) can be expressed as:

\[ \beta_2 = \frac{\lambda_2 D_2}{d_2} \]

### Step 4: Substitute the new conditions into the formula
Now substituting \(D_2\) and \(d_2\) into the equation:

\[ \beta_2 = \frac{\lambda_2 (2D_1)}{\frac{d_1}{2}} = \frac{2 \lambda_2 D_1}{\frac{d_1}{2}} = \frac{2 \lambda_2 D_1 \cdot 2}{d_1} = \frac{4 \lambda_2 D_1}{d_1} \]

### Step 5: Relate the new fringe width to the old fringe width
From the original fringe width formula, we know:

\[ \beta_1 = \frac{\lambda_1 D_1}{d_1} \]

Thus, we can relate \(\beta_2\) to \(\beta_1\):

\[ \beta_2 = \frac{4 \lambda_2 D_1}{d_1} = 4 \cdot \frac{\lambda_2 D_1}{d_1} = 4 \cdot \frac{\lambda_2}{\lambda_1} \cdot \beta_1 \]

### Step 6: Substitute the values
Now substituting the values:

\[ \beta_2 = 4 \cdot \frac{4.0 \times 10^{-7}}{6.4 \times 10^{-7}} \cdot (1 \times 10^{-4}) \]

Calculating \(\frac{4.0}{6.4}\):

\[ \frac{4.0}{6.4} = 0.625 \]

Thus,

\[ \beta_2 = 4 \cdot 0.625 \cdot (1 \times 10^{-4}) = 2.5 \times 10^{-4} \, \text{m} \]

### Final Answer
The value of the new fringe width is:

\[ \beta_2 = 2.5 \times 10^{-4} \, \text{m} \]

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