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 setup where the first diffraction minima occurs at an angle of \( \theta = 30^\circ \) for light of wavelength \( \lambda = 5000 \, \text{Å} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Condition for Minima**:
The condition for the minima in single slit diffraction is given by the formula:
\[ d \sin \theta = n \lambda \]
where \( d \) is the width of the slit, \( \theta \) is the angle of the minima, \( n \) is the order of the minima (for the first minima, \( n = 1 \)), and \( \lambda \) is the wavelength of the light.

2. **Convert Wavelength to Meters**:
The wavelength is given as \( 5000 \, \text{Å} \). We need to convert this to meters:
\[ \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \]

3. **Substitute Known Values**:
For the first minima (\( n = 1 \)), we can rewrite the equation as:
\[ d \sin(30^\circ) = 1 \cdot \lambda \]
Substituting the values we have:
\[ d \sin(30^\circ) = 5 \times 10^{-7} \, \text{m} \]

4. **Calculate \( \sin(30^\circ) \)**:
We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]

5. **Solve for \( d \)**:
Now substituting \( \sin(30^\circ) \) into the equation:
\[ d \cdot \frac{1}{2} = 5 \times 10^{-7} \, \text{m} \]
Multiplying both sides by 2 to solve for \( d \):
\[ d = 2 \cdot 5 \times 10^{-7} \, \text{m} = 10 \times 10^{-7} \, \text{m} \]

6. **Convert to Centimeters**:
To express \( d \) in centimeters, we convert meters to centimeters (1 m = 100 cm):
\[ d = 10 \times 10^{-7} \, \text{m} = 10 \times 10^{-5} \, \text{cm} \]

7. **Final Answer**:
Thus, the width of the slit is:
\[ d = 10 \times 10^{-5} \, \text{cm} \]