Question

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 4 m away from the lens. If wavelength of light used is 5500 Å, then the distance between the first minimum on either side of the central maximum is (`theta` is small and measured in radian) ____________.

विकल्प

• 10^-1 m

• 10^-2 m

• 2 x 10^-2 m

• 2 x 10^-1 m

Answer 1

In Fraunhofer diffraction pattern, slit width is 0.2 mm and screen is at 4 m away from the lens. If wavelength of light used is 5500 Å, then the distance between the first minimum on either side of the central maximum is 2 x 10^-2 m.

Explanation:

Distance of 1^st minima from central maxima

`"y"_(1_"d") = (lambda "D")/"a"`

Distance between two minima on either side of the central maxima is

`2"y"_(1_"d") = (2lambda"D")/"a"`

`= (2 xx 5500 xx 10^-10 xx 4)/( 0.2 xx 10^3)`

`= 0.02  "m"`