In a single slit diffraction of light of wavelength `lambda` by a slit of width e, the size of the central maximum on a screen at a distance b is

Options

A. `2blamda+e`
B. `(2blamda)/(e)`
C. `(2blamda)/(e)+e`
D. `(2blamda)/(e)-e`

Answer

To find the size of the central maximum in a single slit diffraction pattern, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a single slit of width \( e \) and light of wavelength \( \lambda \) passing through it. The screen is located at a distance \( b \) from the slit.

2. **Identify the Position of Minima**:
- The position of the first minima on either side of the central maximum can be found using the formula for minima in single slit diffraction:
\[ e \sin \theta = n \lambda \]
where \( n \) is the order of the minima (for the first minima, \( n = 1 \)).

3. **Small Angle Approximation**:
- For small angles, we can use the approximation \( \sin \theta \approx \tan \theta \approx \theta \). Thus, we can rewrite the equation as:
\[ e \tan \theta = \lambda \]

4. **Relate \( \tan \theta \) to Screen Position**:
- The tangent of the angle \( \theta \) can be expressed in terms of the height \( y \) of the minima on the screen and the distance \( b \) from the slit to the screen:
\[ \tan \theta = \frac{y}{b} \]
- Substituting this into the previous equation gives:
\[ e \frac{y}{b} = \lambda \]

5. **Solve for \( y \)**:
- Rearranging the equation to solve for \( y \) gives:
\[ y = \frac{\lambda b}{e} \]

6. **Calculate the Size of the Central Maximum**:
- The size of the central maximum is defined as the distance between the two first minima on either side of the central peak. Therefore, the total size \( x_0 \) of the central maximum is:
\[ x_0 = 2y \]
- Substituting \( y \) into this equation gives:
\[ x_0 = 2 \left( \frac{\lambda b}{e} \right) = \frac{2\lambda b}{e} \]

7. **Final Expression**:
- The total size of the central maximum on the screen is:
\[ x_0 = \frac{2\lambda b}{e} \]