Light of wavelength `lambda` is incident on a slit of width d and distance between screen and slit is D. Then width of maxima and width of slit will be equal if D is

Options

A. `(2lambda^(2))/d`
B. `d^(2)/(2lambda)`
C. `d/lambda`
D. `(2lambda)/d`

Explanation

To solve the problem, we need to find the distance \( D \) between the slit and the screen such that the width of the maxima (fringe width) is equal to the width of the slit \( d \).

### Step-by-Step Solution:

1. **Understanding Fringe Width**:
The fringe width \( \beta \) in a single-slit diffraction pattern is given by the formula:
\[ \beta = \frac{2\lambda D}{d} \]
where:
- \( \lambda \) is the wavelength of the light,
- \( D \) is the distance from the slit to the screen,
- \( d \) is the width of the slit.

2. **Setting Up the Equation**:
According to the problem, we want the width of the maxima (fringe width \( \beta \)) to be equal to the width of the slit \( d \). Therefore, we can set up the equation:
\[ \beta = d \]
Substituting the expression for \( \beta \):
\[ \frac{2\lambda D}{d} = d \]

3. **Cross-Multiplying**:
To eliminate the fraction, we cross-multiply:
\[ 2\lambda D = d^2 \]

4. **Solving for \( D \)**:
Now, we can solve for \( D \):
\[ D = \frac{d^2}{2\lambda} \]

5. **Final Result**:
Thus, the required distance \( D \) between the slit and the screen, for which the width of the maxima equals the width of the slit, is:
\[ D = \frac{d^2}{2\lambda} \]