In a two slit experiment with monochromatic light, fringes are obtained on a screen placed at some distance from the slits . If the screen is moved by `5xx10^-2m`, towards the slits, the change in fringe width is `3 xx 10^-5`m. If separation between the slits is `10^-3`m, the wavelength of light used is

Options

A. 6000Å
B. 5000Å
C. 3000Å
D. 4500Å

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between fringe width, wavelength, distance, and slit separation
In a two-slit interference experiment, the fringe width (β) 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
- \( d \) = separation between the slits

### Step 2: Determine the change in fringe width
When the screen is moved closer to the slits by a distance \( \Delta D = 5 \times 10^{-2} \, m \), the change in fringe width \( \Delta \beta \) is given as \( 3 \times 10^{-5} \, m \).

### Step 3: Relate the change in fringe width to the change in distance
The change in fringe width can be expressed as:
\[ \Delta \beta = \frac{\lambda \Delta D}{d} \]
Rearranging this gives us:
\[ \lambda = \frac{\Delta \beta \cdot d}{\Delta D} \]

### Step 4: Substitute the known values into the equation
We know:
- \( \Delta \beta = 3 \times 10^{-5} \, m \)
- \( \Delta D = 5 \times 10^{-2} \, m \)
- \( d = 10^{-3} \, m \)

Substituting these values into the equation for \( \lambda \):
\[ \lambda = \frac{(3 \times 10^{-5}) \cdot (10^{-3})}{5 \times 10^{-2}} \]

### Step 5: Calculate the wavelength
Calculating the above expression:
\[ \lambda = \frac{3 \times 10^{-8}}{5 \times 10^{-2}} = 6 \times 10^{-7} \, m \]

### Step 6: Convert the wavelength to Angstroms
Since \( 1 \, \text{m} = 10^{10} \, \text{Å} \):
\[ \lambda = 6 \times 10^{-7} \, m \times 10^{10} \, \text{Å/m} = 6000 \, \text{Å} \]

### Final Answer
The wavelength of light used is \( 6000 \, \text{Å} \).

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