A slit of size `0.15m` is placed at `2.1m` from a screen. On illuminated it by a light of wavelength `5xx10^-5cm`. The width of central maxima will be

Options

A. 70mm
B. 0.14mm
C. 1.4cm
D. 0.14cm

Answer

To find the width of the central maxima in a single-slit diffraction pattern, we can use the formula:

\[ W = \frac{2 \lambda D}{d} \]

where:
- \( W \) is the width of the central maxima,
- \( \lambda \) is the wavelength of the light,
- \( D \) is the distance from the slit to the screen,
- \( d \) is the width of the slit.

### Step-by-Step Solution:

1. **Identify the given values:**
- Slit width \( d = 0.15 \, \text{m} \)
- Distance to the screen \( D = 2.1 \, \text{m} \)
- Wavelength \( \lambda = 5 \times 10^{-5} \, \text{cm} \)

2. **Convert the wavelength from centimeters to meters:**
\[ \lambda = 5 \times 10^{-5} \, \text{cm} = 5 \times 10^{-5} \times 10^{-2} \, \text{m} = 5 \times 10^{-7} \, \text{m} \]

3. **Substitute the values into the formula:**
\[ W = \frac{2 \lambda D}{d} = \frac{2 \times (5 \times 10^{-7}) \times (2.1)}{0.15} \]

4. **Calculate the numerator:**
\[ 2 \times (5 \times 10^{-7}) \times (2.1) = 2 \times 5 \times 2.1 \times 10^{-7} = 21 \times 10^{-7} = 2.1 \times 10^{-6} \, \text{m} \]

5. **Calculate the width of the central maxima:**
\[ W = \frac{2.1 \times 10^{-6}}{0.15} = 1.4 \times 10^{-5} \, \text{m} \]

6. **Convert the width into millimeters:**
\[ W = 1.4 \times 10^{-5} \, \text{m} = 1.4 \, \text{mm} \]

### Final Answer:
The width of the central maxima is \( 1.4 \, \text{mm} \).