A plane wave of wavelength `6250Å` is incident normally on a slit of width `2xx10^(-2)` cm The width of the principal maximum of diffraction pattern on a screen at a distance of 50 cm will be

Options

A. `312.5xx10^(-3) cm`
B. `312.5xx10^(-4)cm`
C. `312 cm`
D. `312.5xx10^(-5) cm`

Explanation

To solve the problem of finding the width of the principal maximum of the diffraction pattern, we can follow these steps:

### Step-by-Step Solution:

1. **Convert Wavelength to Centimeters**:
The given wavelength is \( \lambda = 6250 \, \text{Å} \). We need to convert this to centimeters.
\[ \lambda = 6250 \, \text{Å} = 6250 \times 10^{-10} \, \text{m} = 6250 \times 10^{-8} \, \text{cm} = 6.25 \times 10^{-5} \, \text{cm} \]

2. **Identify Given Values**:
- Width of the slit \( d = 2 \times 10^{-2} \, \text{cm} \)
- Distance to the screen \( D = 50 \, \text{cm} \)

3. **Calculate the Angular Width of the Central Maximum**:
The angular width \( \beta \) of the central maximum is given by:
\[ \beta = 2\theta = \frac{2\lambda}{d} \]
Substituting the values:
\[ \beta = \frac{2 \times 6.25 \times 10^{-5}}{2 \times 10^{-2}} = \frac{6.25 \times 10^{-5}}{10^{-2}} = 6.25 \times 10^{-3} \, \text{radians} \]

4. **Calculate the Linear Width of the Principal Maximum**:
The linear width \( L \) of the principal maximum on the screen is given by:
\[ L = \beta \times D \]
Substituting the values:
\[ L = 6.25 \times 10^{-3} \times 50 = 312.5 \times 10^{-3} \, \text{cm} = 3.125 \, \text{cm} \]

5. **Final Result**:
The width of the principal maximum of the diffraction pattern on the screen is:
\[ L = 3.125 \, \text{cm} \]