Question

The figure below shows the light intensity on a screen 2.3 m behind an aperture. The aperture is illuminated with light of wavelength 680 nm.The aperture is a single slit, what is its width?

Answer 1

Given,

D = 2.3 m ; = 680 nm = 680 x 10^-9 m

Looking at the figure we find that,its a single slit aperture with central maxima's width is double of the side maximas and the intensity is decreasing moving away from the central maxima. The distance between the first minimum and central maximum is

y = 1 cm = 0.01 m.

Now, In the diffraction we have condition for minima in the terms of its slit width.

d sin= m

m = 1 and from small angle approximations: sin = tan = y/D

d y/D =

d = D / y

d = 2.3 x 680 x 10^-9 / y = 1.564 x 10^-6 / 0.01 = 0.00015 m = 0.156 mm

Hence, d = 0.156 mm.