Question

Find the thickness of a soap film that gives constructive second order interference of reflected red light (l = 7000 Å). The index of refraction of the film is 1.33. Assume a parallel beam of incident light directed at 30⁰ to the normal

• A. 4260 Å
• B. 5000 Å
• C. 6200 Å
• D. 6850 Å

Answer 1

Answer: (A)

4260 Å

Answer 2

Within a limited region the film can be considered as a parallel-sided slab of n = 1.33. The optical path difference of the beams reflected at the upper and the lower surfaces is

D = 2nd cos q +$\frac{\lambda}{2}$,

For constructive second-order interference, we have

D = 2l,

or d =$\frac{3\lambda}{4n\cos\theta}$,

where q is the angle of refraction inside the film. From Snell's law, we have

sin q₀ = n sin q,

i.e., sin q =$\frac{1}{n}$sin q₀ =$\frac{1}{2n}$,

or cos q =$\sqrt{1 - \left( \frac{1}{2n} \right)^{2}}$,

which yields d = 4260 Å.