Problem 2. White light is incident on a thin film of Benzene floating on a deep

Question

Problem 2. White light is incident on a thin film of Benzene floating on a deep pool of water indices of refraction are given in the figure below. The color of light observed in reflection is red (650 nm) . The Air nal.o0 Benzene n=1.51 Water n-1.33 a) (5 pts) Explain why top (Air-Benzene) interface, but not from the bottom (Benzene-water) interface the light experiences a phase shift of 180 degrees upon reflection from the (5 pts) In order to decide if the light rays scattered from the top and bottom interfaces are in phase or out of phase, you need to compare the path length difference between rays to the wavelength of the light. For this comparison should you use the wavelength of the light in air, benzene, or water? Explain your answer. b) (5 pts) What is the thinnest possible film of Benzene which would reflect red (650 nm) light? Give your answer in units of nm to three significant digits. c) d) (5 pts) What is the next thickest possible film?

Explanation / Answer

a) phase change in reflection basically depends on the refractive index of the medium in which light is entering. If refractive index of the medium in which light is entering is higher than medium from where light is coming,then 180 degree phase change occurs. It is the case with air-benzene interface.

Opposite of the above statement is when refractive index from where light is coming is higher than index of the entering medium. In this case phase change in reflection wont happen.

I can also be explained throught the relation for reflection vector of the light given by R = (n1 - n2)/(n1 + n2)  

if n2 > n1 then R is negative which shows the change in polarity of electric field vector

if n2< n1 then R is positive which means no change in the polarity of the electric field vector

b)we should use the wavelength of light in benzene..

It is because of the relation path length in the benzene filme = (n-1/2) * (lambda in film) [for in phase]

or path length in the benzene film = (n)*(lambda in film) [for out phase]

therefore it is must to use wavelength in the benzene film

c) Minimum thickness of the film = lambda/(4*n) n is refractive index of benzene.

650 nm/(4*1.51) = 107.615 nm

d) next possbile film is (3*lambda)/(4*n) = 322.847 nm

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