The light of wavelength `6328Å` is incident on a slit of width `0.2mm` perpendicularly, the angular width of central maxima will be

Options

A. (a) `0.36^@`
B. (b) `0.18^@`
C. (c) `0.72^@`
D. (d) `0.09^@`

Explanation

To solve the problem of finding the angular width of the central maxima when light of wavelength \(6328 \, \text{Å}\) is incident on a slit of width \(0.2 \, \text{mm}\), we can follow these steps:

### Step 1: Convert the wavelength to SI units
The wavelength given is \(6328 \, \text{Å}\). We need to convert this into meters.

\[ \lambda = 6328 \, \text{Å} = 6328 \times 10^{-10} \, \text{m} = 6.328 \times 10^{-7} \, \text{m} \]

### Step 2: Convert the slit width to SI units
The slit width is given as \(0.2 \, \text{mm}\). We need to convert this into meters.

\[ A = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m} = 2.0 \times 10^{-4} \, \text{m} \]

### Step 3: Use the formula for angular width of central maxima
The angular half-width \(\theta\) of the central maxima can be calculated using the formula:

\[ \sin \theta = \frac{\lambda}{A} \]

Substituting the values we have:

\[ \sin \theta = \frac{6.328 \times 10^{-7}}{2.0 \times 10^{-4}} \]

### Step 4: Calculate \(\sin \theta\)
Now we perform the calculation:

\[ \sin \theta = \frac{6.328 \times 10^{-7}}{2.0 \times 10^{-4}} = 3.164 \times 10^{-3} \]

### Step 5: Find \(\theta\) using the small angle approximation
For small angles, \(\sin \theta \approx \theta\) (in radians). Therefore:

\[ \theta \approx 3.164 \times 10^{-3} \, \text{radians} \]

### Step 6: Convert \(\theta\) to degrees
To convert radians to degrees, we use the conversion factor:

\[ \theta \, (\text{in degrees}) = \theta \, (\text{in radians}) \times \frac{180}{\pi} \]

Calculating this gives:

\[ \theta \approx 3.164 \times 10^{-3} \times \frac{180}{\pi} \approx 0.181 \, \text{degrees} \]

### Step 7: Calculate the total angular width of the central maxima
The total angular width of the central maxima is given by \(2\theta\):

\[ \text{Total Angular Width} = 2\theta \approx 2 \times 0.181 \approx 0.362 \, \text{degrees} \]

Thus, the final answer is:

\[ \text{Total Angular Width} \approx 0.36 \, \text{degrees} \]

### Summary
The angular width of the central maxima is approximately \(0.36 \, \text{degrees}\).

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