Working on your car you spill oil (index of refraction = 1.55) on the ground int

Question

Working on your car you spill oil (index of refraction = 1.55) on the ground into a puddle of water (n = 1.33). You notice a rainbow pattern appear across the oil slick. Recalling the lessons you learned in physics class, you realize you can calculate where constructive and destructive interference occurs based on the thickness of the oil slick. (Assume that the average wavelength is 578 nm.)

384.37, 434.5, 1153.11

(b) What are the first three thicknesses necessary for destructive interference? These answers are incorrect.

1902,576.5,960.9

Explanation / Answer

here since the reflected rays are out of phase,

the condition for Constructive interference is

t = (m+0.5) * L/2n

where m = 0, 1,2,3 etc

n is refractive index

L is wavelength

so for first thickness

m = 0

t1 = (0 + 0.5) * 578 nm/(2* 1.55)

t1 = 93.22 nm

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seconf thickness

m =1 ,

t2 = (1+ 0.5)*578 *10^-9/(2* 1.55)

t2 = 279.67 nm

for third thickness m= 2

t3 = (2.5* 578 *10^-9)/(2*1.55)

t3 = 466.12 nm

so the three thickness are 93.22 nm, 279.67 nm, 466.12 nm

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Condition for Destructive intrference is

t = mL/2n

for first :

m = 1

t1 = (1 * 578 nm)/(2* 1.55)

t1 = 186.45 nm

t2 = 2* 578nm/(2*.155) = 372.903 nm

t3 = 3*578nm/(2*1.55) = 559.35 nm

so for Destrucitve interferfce the thcknes are

186.45 nm, 372.903nm, 559.35 nm

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