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?

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

b

step-by-step explanation: a third of 24 is 8, 8 times 3 is 24 so 8 - 2 would be 6

b.transmitters send radio waves, and receivers capture radio waves.