If `I_0` is the intensity of the principal maximum in the single slit diffraction pattern. Then what will be its intensity when the slit width is doubled?

Options

A. `4I_0`
B. `2I_0`
C. `(I_0)/(2)`
D. `I_0`

Explanation

To solve the problem, we need to analyze how the intensity of the principal maximum in a single slit diffraction pattern changes when the slit width is doubled.

### Step-by-Step Solution:

1. **Understanding the Intensity in Single Slit Diffraction**:
The intensity \( I \) at the principal maximum in a single slit diffraction pattern is given by the formula:
\[ I = I_0 \left( \frac{\sin \phi}{\phi} \right)^2 \]
where \( \phi = \frac{\pi a \sin \theta}{\lambda} \), \( a \) is the slit width, \( \lambda \) is the wavelength, and \( \theta \) is the angle of diffraction.

2. **Effect of Doubling the Slit Width**:
If the slit width \( a \) is doubled (i.e., \( a' = 2a \)), we need to analyze how this affects the intensity:
- The new angle \( \phi' \) becomes:
\[ \phi' = \frac{\pi (2a) \sin \theta}{\lambda} = 2 \phi \]

3. **Calculating the New Intensity**:
Substitute \( \phi' \) into the intensity formula:
\[ I' = I_0 \left( \frac{\sin(2\phi)}{2\phi} \right)^2 \]
Using the trigonometric identity \( \sin(2\phi) = 2 \sin(\phi) \cos(\phi) \), we can rewrite the intensity as:
\[ I' = I_0 \left( \frac{2 \sin(\phi) \cos(\phi)}{2\phi} \right)^2 = I_0 \left( \frac{\sin(\phi) \cos(\phi)}{\phi} \right)^2 \]

4. **Comparing the New Intensity with the Original Intensity**:
Now, we can relate \( I' \) to \( I_0 \):
\[ I' = I_0 \cdot \left( \frac{\sin(\phi) \cos(\phi)}{\phi} \right)^2 \]
Since \( \cos(\phi) \) is less than or equal to 1, we can see that the intensity \( I' \) will be greater than \( I_0 \) because the area of the slit increases when the width is doubled.

5. **Final Result**:
The intensity of the principal maximum when the slit width is doubled will be:
\[ I' = 4I_0 \]
Therefore, the intensity of the principal maximum when the slit width is doubled is four times the original intensity.