Question

In: Physics

A thin film of oil (n = 1.28) is located on smooth, wet pavement. When viewed...

A thin film of oil (n = 1.28) is located on smooth, wet pavement. When viewed from a direction perpendicular to the pavement, the film reflects most strongly red light at 640 nm and reflects no light at 427 nm. What is the minimum thickness of the oil film?
nm

Solutions

Expert Solution

Refractive index of oil, n = 1.28
?con = 640 nm
?des = 427 nm
Solution:
For constructive interference:
2 t = m ?con / n ...(1)
For destructive interference:
2t = [ m + (1/2) ] ?des / n ...(2)
By equating (1) with (2):
m ?con / n =  [ m + (1/2) ] ?des / n
m ?con =  [ m + (1/2) ] ?des
?con / ?des = 1 + (1/2m)
640 / 427 = 1 + (1/2m)
1.49 = 1 + (1/2m)
0.49 = 1 / 2m
2m = 1/0.49
2m = 5.952
m = 1.02
Hence, thickness (t) = m ?con / 2n
                                = 1* 640 x 10^-9 / 2 * 1.28
                                = 250 x 10^-9
                                = 250 x 10^-9 m
Ans:
Thickness, t = 250 x 10^-9 m (or) 250 nm

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