Question

In Fraunhofer diffraction pattern, slitwidth is 0.5 mm and screen is at 2 m away from the lens. If wavelength of light used is $5500\overset{\circ}{A}$, then the distance between the first minimum on either side of the central maximum is ($\theta$ is small and measured in radian)

• A. 1.1 mm
• B. 2.2 mm
• C. 4.4 mm
• D. 5.5 mm

Answer 1

Answer: (C)

4.4 mm

Answer 2

The distance between the first minima on both sides of the central maximum in Fraunhofer diffraction is given by:

$\Delta x = \frac{2D\lambda}{a}$

Where:

• $D$ is the distance from the slit to the screen.
• $\lambda$ is the wavelength of the light.
• $a$ is the slit width.

Given:

• $a = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m}$
• $D = 2 \, \text{m}$
• $\lambda = 5500 \, \overset{\circ}{A} = 5500 \times 10^{-10} \, \text{m} = 5.5 \times 10^{-7} \, \text{m}$

Substituting these values into the formula:

$\Delta x = \frac{2 \times 2 \, \text{m} \times 5.5 \times 10^{-7} \, \text{m}}{0.5 \times 10^{-3} \, \text{m}} = \frac{4 \times 5.5 \times 10^{-7}}{0.5 \times 10^{-3}} = 4.4 \times 10^{-3} \, \text{m} = 4.4 \, \text{mm}$

Therefore, the distance between the first minimum on either side of the central maximum is 4.4 mm.