Calculate angular width of central maxima if `lamda=6000Å,a=18xx10^(-5)cm`=

Options

A. `20^(@)`
B. `40^(@)`
C. `30^(@)`
D. `260^(@)`

Answer

To calculate the angular width of the central maxima in a single-slit diffraction pattern, we can follow these steps:

### Step 1: Understand the relationship for minima
The condition for the minima in a single-slit diffraction pattern is given by the formula:
\[ a \sin \theta = n \lambda \]
where:
- \( a \) is the slit width,
- \( \theta \) is the angle of the minima,
- \( n \) is the order of the minima (for the first minima, \( n = 1 \)),
- \( \lambda \) is the wavelength of light.

### Step 2: Convert units
Given:
- \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \)
- \( a = 18 \times 10^{-5} \, \text{cm} = 18 \times 10^{-7} \, \text{m} \)

### Step 3: Substitute values into the minima equation
Using \( n = 1 \):
\[ a \sin \theta = \lambda \]
Substituting the values:
\[ 18 \times 10^{-7} \sin \theta = 6 \times 10^{-7} \]

### Step 4: Solve for \( \sin \theta \)
Rearranging gives:
\[ \sin \theta = \frac{6 \times 10^{-7}}{18 \times 10^{-7}} = \frac{1}{3} \]

### Step 5: Find \( \theta \)
Now, we find \( \theta \) using the inverse sine function:
\[ \theta = \sin^{-1}\left(\frac{1}{3}\right) \]

Calculating this gives:
\[ \theta \approx 20^\circ \]

### Step 6: Calculate angular width of central maxima
The angular width of the central maxima is the angle between the first minima on either side of the central peak. Therefore, the total angular width \( W \) is:
\[ W = 2\theta = 2 \times 20^\circ = 40^\circ \]

### Final Answer
The angular width of the central maxima is \( 40^\circ \).

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