Question

For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser pointer (λ = 667 nm) through the slit onto a screen 6.0 m away, a diffraction pattern appears. The bright band in the center of the pattern is 8.0 cm wide. What is the width of the slit?

Answer 1

The wave length of the laser light = 667 nm

the distance between the slit and the screen L = 6.0 m

the width of the bright band 2x = 8.0cm

then x = 4.0cm

The width of the central maximum goes from the first minimum on one side to the first minimum on the other side.
so the first minimum occurs at

sin(theta) = lambda/d

the angle between the center and the first minimum is

sin(theta) ~ tan(theta) = x/L

then x/L = lambda/d

therefore the width of the slit is

d = lambda*L/x

d = (667*10^-9 m)*(6.0 m) / (4.0*10^-2 m) = 1*10^-4 m = 0.01 cm

= 912.2*10-7 m or 912.2*10-4 mm
= 0.0912 mm