The first diffraction minima due to a single slit diffraction is at `theta=30^(@)` for a light of wavelength `5000Å` The width of the slit is

Options

A. `5xx10^(-5)cm`
B. `10xx10^(-5)cm`
C. `2.5xx10^(-5)cm`
D. `1.25xx10^(5)cm`

Answer

To find the width of the slit in a single slit diffraction experiment, we can use the formula for the position of the first diffraction minimum. The condition for the first minimum is given by:

\[ D \sin \theta = N \lambda \]

where:
- \( D \) is the width of the slit,
- \( \theta \) is the angle of the first minimum,
- \( N \) is the order of the minimum (for the first minimum, \( N = 1 \)),
- \( \lambda \) is the wavelength of the light.

### Step-by-Step Solution:

1. **Identify the given values:**
- Wavelength \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} \)
- Angle \( \theta = 30^\circ \)

2. **Use the formula for the first minimum:**
For the first minimum, we can set \( N = 1 \):
\[ D \sin \theta = \lambda \]

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

4. **Substitute the values into the equation:**
\[ D \cdot \frac{1}{2} = 5000 \times 10^{-10} \, \text{m} \]

5. **Solve for \( D \):**
\[ D = 5000 \times 10^{-10} \cdot 2 \]
\[ D = 10000 \times 10^{-10} \, \text{m} \]
\[ D = 1 \times 10^{-6} \, \text{m} \]

6. **Convert to centimeters:**
\[ D = 1 \times 10^{-6} \, \text{m} = 1 \times 10^{-4} \, \text{cm} \]

### Final Answer:
The width of the slit \( D \) is \( 1 \times 10^{-4} \, \text{cm} \).

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