Light of wavelength 600 nm is incident normally on a slit of width 0.2 mm. The angular width of central maxima in the diffraction pattern is

Options

A. `6xx10^(-3) rad`
B. `4xx10^(-3) rad`
C. `2.4xx10^(-3) rad`
D. `4.5xx10^(-3) rad`

Answer

To find the angular width of the central maxima in the diffraction pattern, we can follow these steps:

### Step 1: Convert the Wavelength and Slit Width to Standard Units
- Given wavelength \( \lambda = 600 \, \text{nm} \).
- Convert to meters:
\[ \lambda = 600 \times 10^{-9} \, \text{m} = 6 \times 10^{-7} \, \text{m} \]

- Given slit width \( a = 0.2 \, \text{mm} \).
- Convert to meters:
\[ a = 0.2 \times 10^{-3} \, \text{m} = 2 \times 10^{-4} \, \text{m} \]

### Step 2: Use the Formula for Angular Width of Central Maxima
The angular width \( \theta \) of the central maxima can be calculated using the formula:
\[ \theta = \frac{2\lambda}{a} \]

### Step 3: Substitute the Values into the Formula
Substituting the values we found:
\[ \theta = \frac{2 \times (6 \times 10^{-7})}{2 \times 10^{-4}} \]

### Step 4: Simplify the Expression
- The \( 2 \) in the numerator and denominator cancels out:
\[ \theta = \frac{6 \times 10^{-7}}{10^{-4}} = 6 \times 10^{-3} \, \text{radians} \]

### Step 5: Final Result
The angular width of the central maxima is:
\[ \theta = 6 \times 10^{-3} \, \text{radians} \]