Question

A beam of monochromatic light is incident on a single slit of width 0.640 mm. A diffraction pattern forms on a wall 1.35 m beyond the slit. The distance between the positions of zero intensity on both sides of the central maximum is 2.04 mm. Calculate the wavelength of the light.

Answer 1

positions of zero intensity on both side of central maximum = 2*y = 2*L*tan theta = 2.04 mm

for small angles, tan theta = sin theta

L*sin theta = 2.04/2 = 1.02 mm

Now for single slit diffraction

d*sin theta = m*lambda

sin theta = m*lambda/d

L*m*lambda/d = 1.02 mm

lambda = 1.02*10^-3*d/(m*L)

d = slit width = 0.640 mm

m = 1

L = 1.35 m

So,

lambda = 1.02*10^-3*0.640*10^-3/(1*1.35)

lambda = 4.835*10^-7 = 483.5*10^-9 m

lambda = wavelength = 483.5 nm

Please Upvote.

Comment below if you have any query