Light of wavelength `5000Å` is incident over a slit of width `1 mu m`. The angular width of central maxima will be

Options

A. `30^(@)`
B. `60^(@)`
C. `90^(@)`
D. `120^(@)`

Answer

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

### Step 1: Understand the Problem
We are given:
- Wavelength of light, \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \)
- Width of the slit, \( a = 1 \, \mu m = 1 \times 10^{-6} \, \text{m} \)

We need to find the angular width of the central maxima, which is represented as \( 2\theta \), where \( \theta \) is the angle to the first minimum.

### Step 2: Use the Formula for the First Minimum
The condition for the first minimum in single slit diffraction is given by:
\[ a \sin \theta = n \lambda \]
where \( n = 1 \) for the first minimum.

### Step 3: Substitute the Known Values
Substituting the known values into the equation:
\[ 1 \times 10^{-6} \sin \theta = 1 \times (5 \times 10^{-7}) \]

### Step 4: Solve for \( \sin \theta \)
Rearranging the equation gives:
\[ \sin \theta = \frac{5 \times 10^{-7}}{1 \times 10^{-6}} = 0.5 \]

### Step 5: Find \( \theta \)
Now, we can find \( \theta \):
\[ \theta = \arcsin(0.5) = 30^\circ \]

### Step 6: Calculate the Angular Width of the Central Maxima
The angular width of the central maxima is given by:
\[ \text{Angular width} = 2\theta = 2 \times 30^\circ = 60^\circ \]

### Final Answer
The angular width of the central maxima is \( 60^\circ \).

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