In Young's double slit experiment, the fringe width with light of wavelength 6000 Ã… is 3 mm. The fringe width, when the wavelength of light is changed to 4000 Ã… is

Options

A. 3 mm
B. 1 mm
C. 2 mm
D. 4 mm

Explanation

To solve the problem of finding the fringe width when the wavelength of light is changed from 6000 Å to 4000 Å in Young's double slit experiment, we can 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:
- \( \beta \) = fringe width
- \( \lambda \) = wavelength of light
- \( D \) = distance from the slits to the screen
- \( d \) = distance between the slits

### Step 2: Set up the relationship between the two fringe widths
Let:
- \( \beta_1 \) = fringe width for \( \lambda_1 = 6000 \, \text{Å} \)
- \( \beta_2 \) = fringe width for \( \lambda_2 = 4000 \, \text{Å} \)

From the formula, we can express the fringe widths as:
\[ \beta_1 = \frac{\lambda_1 D}{d} \]
\[ \beta_2 = \frac{\lambda_2 D}{d} \]

### Step 3: Relate the two fringe widths
Dividing the equations for \( \beta_2 \) and \( \beta_1 \):
\[ \frac{\beta_2}{\beta_1} = \frac{\lambda_2}{\lambda_1} \]

### Step 4: Substitute known values
We know:
- \( \beta_1 = 3 \, \text{mm} \)
- \( \lambda_1 = 6000 \, \text{Å} \)
- \( \lambda_2 = 4000 \, \text{Å} \)

Substituting these values into the relationship:
\[ \frac{\beta_2}{3 \, \text{mm}} = \frac{4000 \, \text{Å}}{6000 \, \text{Å}} \]

### Step 5: Solve for \( \beta_2 \)
Calculating the right side:
\[ \frac{4000}{6000} = \frac{2}{3} \]

Now substituting back:
\[ \beta_2 = 3 \, \text{mm} \times \frac{2}{3} = 2 \, \text{mm} \]

### Final Answer
The fringe width when the wavelength of light is changed to 4000 Å is:
\[ \beta_2 = 2 \, \text{mm} \]

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