Question

Light of wavelength 656 nm is incident perpendicularly on a soap film (n = 1.33) suspended in air. What are the (a) least and (b) second least thicknesses of the film for which the reflections from the film undergo fully constructive interference?

Answer 1

For the constructive interference, we should have a phase difference between 1st-surface reflection and
2nd-surface reflection + 2-way trip through film = integer no. of cycles.

Means -

Integer must be > 0.

Phase diff. due to 1st-surface reflection = 0.5 cycles (N1<N2)

Phase diff. due to 2nd-surface reflection = 0 cycles (N2>N1)

Phase diff. due to both reflections = 0.5 cycles.

Now the desired additional phase difference due to 2-way pathlength = integer (incl. 0) + 0.5 cycles.

Wavelength in film lambda2 = lambda1/N2 = 656 / 1.33 = 493.23 nm

(a) So, least value of thickness of the film = lambda2/4 = 493.23 / 4 = 123.31 nm

(b) Second least thickness = (3 * lambda2) / 4 = 369.93 nm