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 558 nm.)

(a) What are the first three thicknesses necessary for constructive interference?
nm
nm
nm

(b) What are the first three thicknesses necessary for destructive interference?
nm
nm
nm

Explanation / Answer

Part A

The constructive interference of thin oil film is

2 noil d = ( m + 1/2) * lambda

==> d = ( m + 1/2) * lambda / ( 2 * noil )

when m =1, lambda = 558 nm ,d = ( 1 + 1/2) * 558 nm / ( 2 * 1.55 ) = 270 nm

  m =2, lambda = 558 nm ,d = ( 2 + 1/2) * 558 nm / ( 2 * 1.55 ) = 450 nm

  m =3, lambda = 558 nm ,d = ( 3 + 1/2) * 558 nm / ( 2 * 1.55 ) = 630 nm

Part B

For destructive interference

2 noil d = m * lambda

==> d = m * lambda / ( 2 * noil)

When m = 1, lamda = 558 nm, d = 1 * 558 nm/ ( 2 * 1.55) = 180 nm

m = 2, lamda = 558 nm, d = 2 * 558 nm/ ( 2 * 1.55) = 360 nm

  m = 3, lamda = 558 nm, d = 3 * 558 nm/ ( 2 * 1.55) = 540 nm

Related Questions

Navigate

• Browse (All)
• Browse W
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.