Question

Light of wavelength 614 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

Answer:

The least and second least thicknesses of the film are 0.115 um and 0.346 um respectively.

Explanation:

Optical path length ===> 2n * t = (m + 0.5) * λ

λ = 614 nm , n = 1.33

Substitute in the parameters in the equation.

∴ 2(1.33) * t = (m + 0.5) * 614

  2.66 * t = 614m + 307

  t = (614m + 307) / 2.66 .............(1)

(a) for m = 0

    t = (614m + 307) / 2.66

    t = (614(0) + 307) / 2.66

    t = 307 / 2.66

    t = 115 nm == 0.115 um

(b) for m = 1

    t = (614(1) + 307) / 2.66

    t = (614 + 307) / 2.66

    t = 921 / 2.66

    t = 346.24 nm = 0.346 um