Question

In a single slit diffraction pattern the distance between the first minimum on the right and the first minimum on the left is 5.2 mm. The screen on which the pattern is displayed is 80 cm from the slit and the wavelength is 5460 $\overset{\circ}{A}$ . Calculate the slit width.

(Ans : 0.168 mm)

Answer 1

The width of the slit is d = 0.4368 mm

Explanation:

The order of the fringe is m = 1,

Distance of first maximum (or minimum) from central fringe is y = 5.2 mm

Distance between screen and slit is D = 80 cm = 800 mm,

The wavelength of incident wave is lambda = 5460∘A

In a single slit diffraction, the slit width is given as

d = mλD  / y

d=    (1 x 800 mm)(5460∘A)  / (5.2 mm)

d = (4368,000) x 10^7

d = 43.68 x 10^5 x 10^-7

d = 43.68 / 100 mm

d = 0.4368 mm

Hence the width of the slit is d = 0.4368 mm

​Also learn more

In young double slit experiment if the distance between two slits is halved and distance between the slits and the screen is doubled then what will be the effect on fringe width ?

https://brainly.in/question/7560303