Question

A 0.50-mm-wide slit is illuminated by light of wavelength 500 nm.

What is the width of the central maximum on a screen 2.0 m behind the slit in mm?

 

Answer 1

We know that

m*lambda = d*sin A

sin A = m*lambda/d

m = 1, for central maximum

lambda = 500 nm

d = 0.50 mm

So,

sin A = 1*500*10^-9/(0.50*10^-3) = 0.001

for small angles

sin A = A = tan A

tan A = y/D

y = D*tan A

y = 2*0.001 = 0.002 m

Now,

width of central maximum = 2y = 2*0.002 = 0.004 m = 4*10^-3 m = 4 mm