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

Answer

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

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have light of wavelength \( \lambda \) passing through a single slit of width \( d \). The distance from the slit to the screen is \( D \). We need to find the condition under which the width of the central maxima on the screen is equal to the width of the slit.

2. **Formula for Width of Central Maxima**:
The width of the central maxima in a single-slit diffraction pattern is given by the formula:
\[ \text{Width of Central Maxima} = \frac{2\lambda D}{d} \]

3. **Setting Up the Equation**:
We want to find \( D \) such that the width of the central maxima equals the width of the slit \( d \). Therefore, we set:
\[ d = \frac{2\lambda D}{d} \]

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

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

### Final Answer:
The distance \( D \) at which the width of the maxima and the width of the slit will be equal is:
\[ D = \frac{d^2}{2\lambda} \]

---