What will be the angular width of central maxima in Fraunhofer diffraction when light of wavelength `6000Å` is used and slit width is `12xx10^-5cm`?

Options

A. (a) `2rad`
B. (b) `3rad`
C. (c) `1rad`
D. (d) `8rad`

Explanation

To find the angular width of the central maxima in Fraunhofer diffraction, we can use the formula:

\[ \text{Angular Width} = 2\theta = \frac{2\lambda}{d} \]

where:
- \(\lambda\) is the wavelength of the light,
- \(d\) is the width of the slit.

### Step 1: Convert the given values to consistent units
The wavelength \(\lambda\) is given as \(6000 \, \text{Å}\) (angstroms), and the slit width \(d\) is given as \(12 \times 10^{-5} \, \text{cm}\).

1. Convert \(\lambda\) from angstroms to meters:
\[ 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \]

2. Convert \(d\) from centimeters to meters:
\[ 12 \times 10^{-5} \, \text{cm} = 12 \times 10^{-5} \times 10^{-2} \, \text{m} = 12 \times 10^{-7} \, \text{m} \]

### Step 2: Substitute the values into the formula
Now, substitute \(\lambda\) and \(d\) into the angular width formula:
\[ 2\theta = \frac{2\lambda}{d} = \frac{2 \times (6 \times 10^{-7} \, \text{m})}{12 \times 10^{-7} \, \text{m}} \]

### Step 3: Simplify the expression
\[ 2\theta = \frac{12 \times 10^{-7}}{12 \times 10^{-7}} = 1 \, \text{radian} \]

### Conclusion
Thus, the angular width of the central maxima is:
\[ \text{Angular Width} = 1 \, \text{radian} \]

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