Angular width of central maxima is `pi//2`, when a slit of width 'a' is illuminated by a light of wavelength `7000 Å` then a =

Options

A. `9xx10^(-9)m`
B. `8.0xx10^(-7)m`
C. `9xx10^(-7)m`
D. `9.8xx10^(-7)m`

Explanation

To solve the problem, we need to find the width of the slit 'a' given that the angular width of the central maximum is \(\frac{\pi}{2}\) and the wavelength of light is \(7000 \, \text{Å}\).

### Step-by-Step Solution:

1. **Understand the Angular Width of the Central Maximum**:
The angular width of the central maximum is given as \(\frac{\pi}{2}\). This means that the angle from the center of the central maximum to the first minimum on either side is \(\frac{\pi}{4}\).

2. **Use the Condition for the First Minimum**:
The condition for the first minimum in single-slit diffraction is given by:
\[ a \sin \theta = n \lambda \]
where \(n\) is the order of the minimum (for the first minimum, \(n = 1\)), \(a\) is the slit width, \(\lambda\) is the wavelength, and \(\theta\) is the angle to the first minimum.

3. **Substituting the Values**:
Since we have determined that the angle \(\theta\) to the first minimum is \(\frac{\pi}{4}\) (or \(45^\circ\)), we can substitute this into the equation:
\[ a \sin\left(\frac{\pi}{4}\right) = 1 \cdot \lambda \]
The sine of \(45^\circ\) is \(\frac{1}{\sqrt{2}}\), so we can write:
\[ a \cdot \frac{1}{\sqrt{2}} = \lambda \]

4. **Substituting the Wavelength**:
The wavelength \(\lambda\) is given as \(7000 \, \text{Å}\), which can be converted to meters:
\[ \lambda = 7000 \, \text{Å} = 7000 \times 10^{-10} \, \text{m} = 7 \times 10^{-7} \, \text{m} \]

5. **Solving for Slit Width 'a'**:
Now we can rearrange the equation to find \(a\):
\[ a = \lambda \cdot \sqrt{2} \]
Substituting the value of \(\lambda\):
\[ a = 7 \times 10^{-7} \cdot \sqrt{2} \]

6. **Calculating the Final Value**:
We can calculate \(a\) as follows:
\[ a = 7 \times 10^{-7} \cdot 1.414 \approx 9.899 \times 10^{-7} \, \text{m} \]
Rounding this gives:
\[ a \approx 9.8 \times 10^{-7} \, \text{m} \]

### Final Answer:
The width of the slit \(a\) is approximately \(9.8 \times 10^{-7} \, \text{m}\).