What will be the angular width of central maxima in Franuhoffer diffraction when light of wavelength 12000 nm falls on slit of width 5mm ?

Options

A. 5mm
B. 4.8 mm
C. 4.9 mm
D. 2 mm

Explanation

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

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

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

### Step 1: Convert the given values to standard units
- Wavelength \(\lambda = 12000 \, \text{nm} = 12000 \times 10^{-9} \, \text{m} = 1.2 \times 10^{-5} \, \text{m}\)
- Slit width \(d = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m}\)

### Step 2: Substitute the values into the formula
Now we can substitute the values of \(\lambda\) and \(d\) into the formula for angular width:

\[ \text{Angular width} = \frac{2 \times (1.2 \times 10^{-5})}{5 \times 10^{-3}} \]

### Step 3: Calculate the angular width
Calculating the above expression:

\[ \text{Angular width} = \frac{2.4 \times 10^{-5}}{5 \times 10^{-3}} = \frac{2.4}{5} \times 10^{-2} = 0.48 \times 10^{-2} \, \text{radians} \]

### Step 4: Convert to millimeters for clarity
To express the angular width in a more understandable unit, we can convert it to millimeters:

\[ 0.48 \times 10^{-2} \, \text{radians} = 4.8 \times 10^{-3} \, \text{radians} = 4.8 \, \text{mm} \]

### Final Answer
Thus, the angular width of the central maxima is:

\[ \text{Angular width} = 4.8 \, \text{mm} \]

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