A soap film ( n =1.33) is 753nm thick. White light strikes the film at normal in

Question

A soap film (n=1.33) is 753nm thick. White light strikes the film at normal incidence.

Part A

What visible wavelengths will be constructively reflected if the film is surrounded by air on both sides?

If there is more than one answer, separate them by a comma.

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A soap film (n=1.33) is 753nm thick. White light strikes the film at normal incidence.

Part A

What visible wavelengths will be constructively reflected if the film is surrounded by air on both sides?

If there is more than one answer, separate them by a comma.

SubmitMy AnswersGive Up

Explanation / Answer

For a reflection maximum we want phase difference between 1st-surface reflection and 2nd-surface reflection + 2-way trip through film = integer no. of cycles.
N1 = N3 = 1, N2 = 1.33
Phase diff. due to 1st-surface reflection = 0.5 cycles
Phase diff. due to 2nd-surface reflection = 0 cycles
Phase diff. due to both reflections = 0.5 cycles
Desired add'l. phase diff. due to pathlength phiD = integer + 0.5 cycles
Air equiv. 2-way path wavelength PLambda1 = 2T*N2/N1 = 2002.98 nm
First few lambda (air) values:
Lambda/2 phase shift in 2-way path ==> lambda = 4T/(1,3,5...)
for n =1

lambda1 = 4*2002.98/2 = 4005.96 nm

lambda2 = 4*2002.98/2*3 = 1335.32 nm

lambda3 = 4*2002.98/2*5 = 801.192 nm

lambda4 = 4*2002.98/2*7 = 572.28 nm

lambda5 = 4*2002.98/2*9 = 445.1067 nm

lambda6 = 364.178 nm

In the visible range( 700 nm - 400 nm) we have 572.28 nm and 445.1067 nm

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