Light of wavelengths 500nm is incident normally on a film of water 0.94x10^-6 m

Question

Light of wavelengths 500nm is incident normally on a film of water 0.94x10^-6 m thick. The index of refration of water is 1.33

a)how many wave lengths are contained in the distance 2L?
b)What is the phase difference between the waves reflected from the topof the film and the one reflected from the bottom after it has travelled this distance? Do you get constructive of destructive interference?
c)How much thicker would the film need to be in order to get the opposite type of interference to that in b???

Explanation / Answer

a) The wavelength of light in water will be,

= 0/n = 500/1.33 = 375 nm

In the distance of 2L the number of wavelengths will be,

N = 2L/ = 2*0.94e-6/375e-9 5 wavelengths

b) The path difference here is x=5. The condition for constructive interference is that the path difference must be an integer multiple of the wavelength.

We get a constructive interference.

c) For a destructive inteference, x has to be an "integer plus half" multiple of the wavelength. Therefore if we increase or decrease x by /2 we will get destructive interference.

The condition for destructive interference will be obtained if 2L is changed by /2 or if L is changed by /4.

/4 = 375/4 = 93.75 nm

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