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. 3200 Ã…
B. `6.5xx10^(-4)mm`
C. 1.3 micron
D. `2.6xx10^(-4)cm`

Explanation

To solve the problem, we need to find the width of the slit (denoted as 'd') using the given information about the first minima for red light. Here's the step-by-step solution:

### Step 1: Write down the given data
- Wavelength of red light, \( \lambda = 6500 \, \text{Å} = 6500 \times 10^{-10} \, \text{m} \)
- Angle for the first minima, \( \theta = 30^\circ \)

### Step 2: Use the formula for the first minima in single-slit diffraction
The condition for the first minima in single-slit diffraction is given by:
\[ d \sin \theta = \lambda \]
Where:
- \( d \) is the width of the slit
- \( \theta \) is the angle of the first minima
- \( \lambda \) is the wavelength of the light

### Step 3: Rearrange the formula to solve for \( d \)
Rearranging the formula gives:
\[ d = \frac{\lambda}{\sin \theta} \]

### Step 4: Substitute the values into the equation
Now substitute the values of \( \lambda \) and \( \theta \):
\[ d = \frac{6500 \times 10^{-10} \, \text{m}}{\sin(30^\circ)} \]

### Step 5: Calculate \( \sin(30^\circ) \)
We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]

### Step 6: Substitute \( \sin(30^\circ) \) into the equation
Now substituting this value back into the equation:
\[ d = \frac{6500 \times 10^{-10} \, \text{m}}{\frac{1}{2}} = 6500 \times 10^{-10} \, \text{m} \times 2 \]
\[ d = 13000 \times 10^{-10} \, \text{m} \]

### Step 7: Convert to microns
To convert meters to microns (where \( 1 \, \text{micron} = 10^{-6} \, \text{m} \)):
\[ d = 13000 \times 10^{-10} \, \text{m} = 13000 \times 10^{-4} \, \text{micron} = 1.3 \, \text{micron} \]

### Final Answer
The width of the slit \( d \) is \( 1.3 \, \text{micron} \).

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