A slit of width is illuminated by white light. For red light `(lambda=6500Å)`, the first minima is obtained at `theta=30^@`. Then the value of will be

Options

A. (a) `3250Å`
B. (b) `6.5xx10^-4mm`
C. (c) `1.24`
D. (d) `2.6xx10^-4mm`

Explanation

To solve the problem, we need to find the width of the slit (denoted as 'a') given that the first minima for red light (with a wavelength of \( \lambda = 6500 \, \text{Å} \)) occurs at an angle \( \theta = 30^\circ \).

### Step-by-Step Solution:

1. **Understand the Condition for Minima:**
The condition for the first minima in single-slit diffraction is given by the formula:
\[ a \sin \theta = m \lambda \]
where \( m \) is the order of the minima (for the first minima, \( m = 1 \)).

2. **Rearranging the Formula:**
For the first minima (\( m = 1 \)), we can rearrange the formula to find the slit width \( a \):
\[ a = \frac{\lambda}{\sin \theta} \]

3. **Convert Wavelength to Meters:**
The wavelength \( \lambda = 6500 \, \text{Å} \) needs to be converted to meters:
\[ 6500 \, \text{Å} = 6500 \times 10^{-10} \, \text{m} = 6.5 \times 10^{-7} \, \text{m} \]

4. **Convert Angle to Radians:**
The angle \( \theta = 30^\circ \) must be converted to radians:
\[ \theta = 30^\circ \times \frac{\pi}{180^\circ} = \frac{\pi}{6} \, \text{radians} \]

5. **Calculate \( \sin \theta \):**
Now, calculate \( \sin(30^\circ) \):
\[ \sin(30^\circ) = \frac{1}{2} \]

6. **Substituting Values into the Formula:**
Now substitute the values of \( \lambda \) and \( \sin \theta \) into the formula for \( a \):
\[ a = \frac{6.5 \times 10^{-7}}{\frac{1}{2}} = 6.5 \times 10^{-7} \times 2 = 1.3 \times 10^{-6} \, \text{m} \]

7. **Convert to Micrometers:**
Convert the slit width from meters to micrometers:
\[ a = 1.3 \times 10^{-6} \, \text{m} = 1.3 \, \mu m \]

### Final Answer:
The width of the slit \( a \) is \( 1.3 \, \mu m \).

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