A thin film of oil (n = 1.29) is located on smooth, wet pavement. When viewed fr

Question

A thin film of oil (n = 1.29) 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 548 nm. What is the minimum thickness of the oil film? Your response differs from the correct answer by more than 10%. Double check your calculations, nm (b) Let mi correspond to the order of the constructive interference and m_2 to the order of destructive interference. Obtain a relationship between m_1 and m_2 that is consistent with the given data. This answer has not been graded yet.

Explanation / Answer

Let the refractive index of the oil is n = 1.29

Let the thickness of the film be t

The extra distance the rays travel is 2 t

Then the optical path length is given by x = 2 t n

For red light the constructive interference condition is 2 * t * n = m * r ………………. ( 1 )

For no light the destructive interference condition is 2 * t * n = ( m – ½ ) * 0 …………….. ( 2 )

We assume that order is same

Then ( m – ½ ) 0 – m r = 0 ……………… ( 3 )

We get m = 0 / [ 2 ( 0 – r ) ]

                 = 548 * 10-9 / [ 2 * ( 548 – 640 ) * 10-9 ]

                 = 548 / 184

                = 2.978

On substituting the value of the m in ( 1 ) , we get

       2 * t * 1.29 = 2.978 * 640 * 10-9

                       t = 1906 nm / 2.58 = 738.79 nm

so the thickness of the film is t = 738.79 nm

b)Let the order are not same and an order m1 corresponds to constructive interference and m2 corresponds to destructive interference

then ( m2 – ½ ) 0 – ( m1 * r ) = 0

         ( m2 – ½ ) 0 = m1 * r

So that    m2 – 1/2 = m1 * ( r / 0 )

              m2 – 1/2 = m1 * ( 640 nm / 548 nm )

                 m2 – 1/2 = 1.167 m1

    we get 2.33 m1 = 2 m2 – 1

we can write    m1 = ( 2 m2 – 1 ) / 2.33

If we take m1 = m2 , we can get m1 = m2 = m = 2.978

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