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A very thin film of oil (n-1.25) sits on glass (n -1.53), What is the phase shift of the light reflected from (a) the top surface of the film? (b)the bottom surface of the film? (c) What is the thinnest film that produces a strong cancellation on the reflection at 600 nm?

User PW Kad
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1 Answer

3 votes

Answer:

240 nm

Step-by-step explanation:

In the interference in thin films , light reflected from upper surface and light reflected from the lower surface of oil are made to interfere each other to form fringes due to path difference of 2μt .

a ) An additional phase shift of 180 ° occurs on the upper surface because light is going from optically rarer medium to denser medium .

b ) There is no such change on reflection from lower surface as light is going from denser to rarer medium ( air ).

c ) For destructive interference the path difference which is equal to 2 μ t should be equal to n λ where n is any number

So for given condition

2 μ t = n λ

For minimum thickness

2 μ t = λ ( n =1 )

t = λ / 2μ

= 600 x 10⁻⁹ /( 2 x 1.25 )

= 240 nm

User Seddikomar
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