Kaplan FL 4 #19 Wavelength/Interference

[Soccerdoc11]

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19. One application of the interferometer is the determination of the index of refraction of a gas sample. If the wavelength of light in vacuum is λ0, then how many wavelengths will fit in a distance d in a medium with an index of refraction of n? (The speed of light in a medium with an index of refraction n is given by v = c/n.)

nλ0/d

nd/λ0

λ0/nd

d/nλ0

B is correct.

Anyways, I hate their explanation, can anyone else derive this in a clearer way? Thanks in advance..

 

[HawkeyePostOp]

c = f*lambda
frequency of light (f) doesn't change in new medium

but speed inchanges to c/n

so lambda changes to (lambda_O)/n

the number of segments in the given distance d would be d/(lambda_O/n) = nd/lambda_O

 

[Soccerdoc11]

Thanks... I finally got that "Aha" moment, makes complete sense