A parallel beam of monochromatic light of wavelength `5000Å` is incident normally on a single narrow slit of width `0.001mm`. The light is focused by a convex lens on a screen placed on the focal plane. The first minimum will be formed for the angle of diffraction equal to

Options

A. `0^@`
B. `15^@`
C. `30^@`
D. `60^@`

Answer

To solve the problem of finding the angle of diffraction for the first minimum formed by a single narrow slit, we will follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Wavelength of light, \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \)
- Width of the slit, \( d = 0.001 \, \text{mm} = 0.001 \times 10^{-3} \, \text{m} = 10^{-6} \, \text{m} \)

2. **Use the Formula for Minima in Single Slit Diffraction**:
The condition for the first minimum in single slit diffraction is given by:
\[ d \sin \theta = n \lambda \]
where \( n \) is the order of the minimum. For the first minimum, \( n = 1 \).

3. **Substitute the Values**:
For the first minimum, we have:
\[ d \sin \theta = 1 \cdot \lambda \]
Substituting the values of \( d \) and \( \lambda \):
\[ 10^{-6} \sin \theta = 5 \times 10^{-7} \]

4. **Solve for \( \sin \theta \)**:
Rearranging the equation gives:
\[ \sin \theta = \frac{5 \times 10^{-7}}{10^{-6}} = 0.5 \]

5. **Find the Angle \( \theta \)**:
To find \( \theta \), we take the inverse sine:
\[ \theta = \sin^{-1}(0.5) \]
This gives:
\[ \theta = 30^\circ \]

6. **Conclusion**:
The angle of diffraction for the first minimum is \( 30^\circ \).

### Final Answer:
The first minimum will be formed for the angle of diffraction equal to \( 30^\circ \).

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