Question 289: Waves Optics

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 $5500\mathop A\limits^o$, then the distance between the first minimum on either side of the central maximum is ( $	heta$ is small and measured in radian)

Accepted Answer

In a Fraunhofer diffraction pattern, the slit width is 0.5 mm, and the screen is placed 2 meters away from the lens. The light used has a wavelength of $5500 \, 	ext{ÃÂ}$. To find the distance between the first minima on either side of the central maximum, we use the following formulas (assuming $	heta$ is small and measured in radians):
The distance of the first minimum from the central maximum is given by:
$ y_{1d} = \frac{\lambda D}{a} $
Where:
$ \lambda = 5500 	imes 10^{-10} \, 	ext{m} $ (wavelength converted from angstroms to meters)
$ D = 2 \, 	ext{m} $ (distance to the screen)
$ a = 0.5 	imes 10^{-3} \, 	ext{m} $ (slit width converted from millimeters to meters)
The distance between the first minima on either side of the central maximum is:
$ 2 y_{1d} = \frac{2 \lambda D}{a} = \frac{2 	imes 5500 	imes 10^{-10} 	imes 2}{0.5 	imes 10^{-3}} = 4.4 \, 	ext{mm} $
Therefore, the distance between the first minima on either side of the central maximum is 4.4 mm.

Answer

1.1 mm

Answer

2.2 mm

Answer

4.4 mm

Answer

5.5 mm

Waves Optics

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\mathop A\limits^o, then the distance between the first minimum on either side of the central maximum is ( \theta is small and measured in radian)

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