Question

a double slit interference pattern is shown below

A double slit interference pattern is shown below. Assume that the angle 6pi in the figure corresponds to a distance of 4.80 mm from the central peak of the pattern, that the wavelength o the light is that of the HeNe laser. 633 nm, and that the interference pattern is being formed on a screen 2.00 m from the slits. Determine values for the slit separation d and slit width a, both in mm. Your values need not be exact but they should be reasonably close to the actual values.

Answer 1

for double slit inteference

the position of minima is given by y = m*lambda*R/2d

m = 1 , 3, 5 ....

for first order minima m = 1

given y = 4.8*10^-3 m

R = distance of screen from slit = 2 m

lambda = wavelength = 633*10^-9 m

d = slits seperation

4.8*10^-3 = 1*633*10^-9*2/(2*d)

d = 1.318*10^-4 m

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for single slit diffraction

y = m*lambda/a

here m = 6 the minima

a = slit width

4.8*10^-3 = 6*633*10^-9/a

a = 7.9125*10^-4 m