Light of wavelength `lambda` is incident on a single sht of width. 'a' and the distance between slit and screen is 'D. In diffraction pattern, if slit width is equal to the width of the central maximum then 'O is equal to

Options

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

Explanation

To solve the problem, we need to analyze the diffraction pattern created by a single slit of width 'a' when light of wavelength 'λ' is incident on it. The goal is to find the distance 'O' from the center of the central maximum to the first minimum in the diffraction pattern, given that the width of the slit is equal to the width of the central maximum.

### Step-by-step Solution:

1. **Understanding the Diffraction Pattern**:
- When light passes through a single slit, it creates a diffraction pattern on a screen placed at a distance 'D' from the slit.
- The width of the central maximum is determined by the position of the first minimum on either side of the central maximum.

2. **Position of the First Minimum**:
- The position of the first minimum in a single-slit diffraction pattern is given by the formula:
\[ y_1 = \frac{\lambda D}{a} \]
- Here, \(y_1\) is the distance from the center of the central maximum to the first minimum, \(λ\) is the wavelength of light, \(D\) is the distance from the slit to the screen, and \(a\) is the width of the slit.

3. **Width of the Central Maximum**:
- The width of the central maximum is defined as the distance between the first minimum on one side and the first minimum on the other side. Therefore, the total width of the central maximum is:
\[ W = 2y_1 = 2 \left(\frac{\lambda D}{a}\right) = \frac{2\lambda D}{a} \]

4. **Setting the Width of the Slit Equal to the Width of the Central Maximum**:
- According to the problem, the width of the slit \(a\) is equal to the width of the central maximum \(W\):
\[ a = \frac{2\lambda D}{a} \]

5. **Solving for D**:
- Rearranging the equation gives:
\[ a^2 = 2\lambda D \]
- Therefore, we can express \(D\) in terms of \(a\) and \(λ\):
\[ D = \frac{a^2}{2\lambda} \]

6. **Finding the Value of O**:
- The distance \(O\) from the center of the central maximum to the first minimum is:
\[ O = y_1 = \frac{\lambda D}{a} \]
- Substituting the expression for \(D\):
\[ O = \frac{\lambda}{a} \cdot \frac{a^2}{2\lambda} = \frac{a}{2} \]

### Final Answer:
Thus, the distance \(O\) is equal to:
\[ O = \frac{a}{2} \]