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

Options

A. (a) `0.20cm`
B. (b) `0.15cm`
C. (c) `0.10cm`
D. (d) `0.25cm`

Explanation

To solve the problem, we need to find the distance of the first dark band of the diffraction pattern from the center of the screen. We will use the formula for the position of dark fringes in a single-slit diffraction pattern.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Width of the aperture (d) = 0.02 cm = 0.02 x 10^-2 m = 2 x 10^-4 m
- Focal length of the lens (f) = 60 cm = 0.60 m
- Wavelength (λ) = 5 x 10^-5 cm = 5 x 10^-7 m

2. **Determine the Distance to the Screen (D):**
- Since the aperture is placed immediately in front of the lens, the distance from the aperture to the screen (D) is equal to the focal length of the lens.
- Therefore, D = f = 0.60 m.

3. **Calculate the Fringe Width (β):**
- The formula for the fringe width (β) in a single-slit diffraction pattern is given by:
\[ \beta = \frac{\lambda D}{d} \]
- Substituting the values:
\[ \beta = \frac{(5 \times 10^{-7} \text{ m}) \times (0.60 \text{ m})}{(2 \times 10^{-4} \text{ m})} \]

4. **Perform the Calculation:**
- Calculate the numerator:
\[ 5 \times 10^{-7} \times 0.60 = 3 \times 10^{-7} \text{ m} \]
- Now, calculate β:
\[ \beta = \frac{3 \times 10^{-7}}{2 \times 10^{-4}} = 1.5 \times 10^{-3} \text{ m} = 0.0015 \text{ m} = 0.15 \text{ cm} \]

5. **Find the Position of the First Dark Band:**
- The position of the first dark band (y) from the center is given by:
\[ y = \frac{\beta}{2} \]
- Thus:
\[ y = \frac{0.15 \text{ cm}}{2} = 0.075 \text{ cm} \]

### Final Answer:
The distance of the first dark band of the diffraction pattern from the center of the screen is **0.075 cm**.