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?

Question

Physics

Janiah Conley

7 September, 08:14

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?

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Answers (1)

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Jonas Duarte · Answer 1

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