Question

3. Interference: Consider the double slit experiment conducted in lab. You now have a new double slit slide to analyze. On the wall 1.75m away from the slits, the first minima occurs 0.65 mm to the side of the central maxima when using the red laser (wavelength=633 nm).

a) Find the slit spacing.

b) If the fourth maximum (from the central maximum) is missing, what is the slit width? Be sure to justify your answer with the appropriate equation(s).

c) What is the double slit phase at the third double slit minima?

d) What is the single slit phase at the third double slit minima?

e) What is the combined intensity at the third double slit minima?

Answer 1

a] For 1st minima due to interference by double-slit,

where d is the slit spacing, x is the distance from the central maximum to 1st minima and D is the distance between the screen and the slits.

=>

=> d = 8.521 x 10^-4 m = 0.8521 mm

b] For mth maxima due to double-slit interference,

so, for 4th maxima,

substituting the values of d and D gives,

x = 5.2 x 10^-3 = 5.2 mm

this is the position of the 4th maxima. Since this order is missing, it must be overlapping with the 1st minima due to diffraction (the slits are of finite width 'a')

so, for 1st minima of single slit,

=>

=> a = 2.13 x 10^-4 m = 0.213 mm

this is the width of each slit.

c] Double slit phase for small-angle approximation is given by:

but for 3rd minima,

therefore,

this is the double-slit phase for 3rd minima.

d] Single slit phase for small-angle approximation is given by:

but for 3rd double-slit minima,

therefore, the single slit phase for 3rd double-slit minima will be:

e] Combined intensity at the third double slit minima will be:

=> .