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. (a) `0^@`
B. (b) `15^@`
C. (c) `30^@`
D. (d) `60^@`

Explanation

To solve the problem, we need to determine the angle of diffraction at which the first minimum occurs when monochromatic light passes through a single narrow slit.

### Step-by-Step Solution:

1. **Identify the 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, \( D = 0.001 \, \text{mm} = 0.001 \times 10^{-3} \, \text{m} = 1 \times 10^{-6} \, \text{m} \)

2. **Use the Condition for the First Minimum:**
The condition for the first minimum in single-slit diffraction is given by:
\[ D \sin \theta = \lambda \]

3. **Rearranging the Equation:**
To find \( \sin \theta \), we rearrange the equation:
\[ \sin \theta = \frac{\lambda}{D} \]

4. **Substituting the Values:**
Now, substitute the values of \( \lambda \) and \( D \):
\[ \sin \theta = \frac{5 \times 10^{-7}}{1 \times 10^{-6}} = 0.5 \]

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

6. **Calculating the Angle:**
From trigonometric values, we know:
\[ \theta = 30^\circ \]

### Final Answer:
The angle of diffraction at which the first minimum will be formed is \( 30^\circ \).

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