Patrouille aérienne de contrôle de la pollution pétrolière 4:
Franges d'interférence des hydrocarbures sur l'eau: Exercices supplémentaires (en anglais)

Airborne Oil Pollution Patrol 4: Interference Fringes of Oil on Water: Supplement

Scenario

We want to get a deeper understanding of the interference fringes observed with microwave radiometry. In the graph, the geometric difference of the path of rays Δs of the waves is:

Δs= n oil ( BC ¯ + CD ¯ ) BE ¯

Using trigonometric relations and Snell’s Law, this becomes:

Δs=2d n oil 2 sin 2 γ

Due to this geometric path difference, there is a phase difference Δϕ of the waves:

Δϕ= 2π λ Δs

At point C, the wave undergoes a reflection at a medium with higher refractive index (i.e., the water). Hence, there is a phase jump of π of the reflected wave. The phase difference of the waves becomes:

Δϕ= 2π λ Δsπ

where the term -π is from the phase jump at point C. There is constructive interference with Δϕ=πm and destructive interference with Δϕ=π(2m+1) , where m is an integer number.

Zoom Sign
Interference of oil on water
An oil layer with thickness d on the water surface, and microwaves with a path of rays directed towards the detector of the microwave radiometer.

Exercises

This is what students can do (continued from Worksheet Marine Pollution C02-03):

  1. Using trigonometric relations and Snell’s Law, show that the geometric path difference Δs of the waves holds as given above, depending on the oil layer thickness d, the refractive index n of the oil, and the incidence angle α.
  2. Please calculate the phase difference of waves for an oil layer thickness from 0 to 5 mm in steps of 0.25 mm with the following data: f=34 GHz, noil=1.41, γ=50°. Take also into account the phase jump of the wave reflected at point C. Verify the correctness of the position of maxima and minima of the brightness temperature curve shown on page 2 of the supplement on Microwave Radiometers.

Solution to exercises

  1. Distances AB ¯ , BC ¯ , and CD ¯ are identical. Therefore:                         n oil ( BC ¯ + CD ¯ )= 2 n oil d cosβ

    The following relations hold:                                        sinγ= BE ¯ BD ¯ tanβ= BD ¯ 2d tanβ= sinβ cosβ

    With sinγ / sinβ = n oil one obtains:                         Δs= 2nd cosβ 2nd sin 2 β cosβ =2ndcosβ=2d n oil 2 sin 2 γ

Materials

Time needed

Procedure

Background Information