A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60cm. The aperture is illuminated normally by a parallel beam of wavelength `5xx10^(-5)` cm. The distance of the first dark band of the diffraction pattern from the centre of the screen is

Options

A. `0.10 cm`
B. `0.25 cm`
C. `0.20 cm`
D. `0.15 cm`

Explanation

To solve the problem step by step, we will use the formula for the position of the first dark band in a single-slit diffraction pattern.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Width of the aperture (d) = 0.02 cm
- Focal length of the lens (D) = 60 cm
- Wavelength of light (λ) = 5 × 10^(-5) cm

2. **Use the Formula for the Position of the First Minimum:**
The position of the first dark band (y) in the diffraction pattern can be calculated using the formula:
\[ y = \frac{n \cdot \lambda \cdot D}{d} \]
where:
- n = 1 (for the first minimum)
- λ = wavelength
- D = distance from the aperture to the screen (focal length of the lens)
- d = width of the aperture

3. **Substitute the Values into the Formula:**
\[ y = \frac{1 \cdot (5 \times 10^{-5}) \cdot 60}{0.02} \]

4. **Calculate the Numerator:**
- First, calculate \( 5 \times 60 = 300 \).
- So, the numerator becomes \( 300 \times 10^{-5} \).

5. **Calculate the Denominator:**
- The denominator is \( 0.02 \) cm.

6. **Perform the Division:**
\[ y = \frac{300 \times 10^{-5}}{0.02} \]
- Convert \( 0.02 \) to a fraction: \( 0.02 = \frac{2}{100} = \frac{1}{50} \).
- Therefore, \( y = 300 \times 10^{-5} \times 50 = 15000 \times 10^{-5} \).

7. **Convert to Standard Form:**
- \( 15000 \times 10^{-5} = 15 \times 10^{3} \times 10^{-5} = 15 \times 10^{-2} \) cm.

8. **Final Calculation:**
\[ y = 0.15 \text{ cm} \]

### Conclusion:
The distance of the first dark band of the diffraction pattern from the center of the screen is **0.15 cm**.