Question

A helium-neon laser (λ=633nm) illuminates a single slit and is observed on a screen 1.50 mbehind the slit. The distance between the first and second minima in the diffraction pattern is 4.95 mm

What is the width (in mm) of the slit?

Answer 1

The expression for the position of the m-th minimum from a single slit diffraction pattern is given as -

y = mλD/d

where
m = the order of minimum
y = distance on the screen of the minimum from central axis
λ = wavelength
D = screen-to-slit distance
d = slit width

Therefore, the difference in y values between the 1st and 2nd minima is -

y₂ - y₁ = [2 x (633x10^-9) x 1.50 / d] - [1 x (633x10^-9) x 1.50 / d]
= 633x10^-9 x 1.50 / d
= 949.5x10^-9 / d

again we have -

y₂ - y₁ = = 4.95 mm = 0.00495 m

therefore -

949.5x10^-9 / d = 0.00495

=> d = (949.5 x 10^-9) / 0.00495 = 1.92 x 10^-4 m = 0.192 x 10^-3 m = 0.192 mm