Anonymous739
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A 500-nm wavelength light in vacuum illuminates a soap film with an index of refraction of 1.33. Air (n=1.00) is on both sides of the film. If the light strikes the film nearly perpindicularly, what is the smallest film thickness such that the film appears bright?
ANS --> 94.0

Please show your work as to how to end up with this answer.

Respuesta :

Answer:

Wavelength of light in film (let y = lambda)

y = 500 nm / (4/3) = 375 nm    

There will be a phase change at the air/film interface (not the other side)

S = 4 t       thickness of film = S/4 where S equals 1 wavelength

This is because of the phase change at one surface

375 nm = 4 * t

t = 93.8 nm

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