The wavelength given in the problem is the wavelength in air. Remember that you

Question

The wavelength given in the problem is the wavelength in air. Remember that you need to adjust for the fact that you want the wavelength in the film.

What are the three smallest non–zero thicknesses of soapy water (n = 1.33) on Plexiglass (n = 1.51) if it appears blue green (constructively reflecting 470–nm light) when illuminated perpendicularly by white light? Explicitly show how you follow the steps in Problem Solving Strategies for Wave Optics. Enter your answers of the thicknesses in ascending order. t1 = t2 = t3 =

Explanation / Answer

Strategy and Concept:

Note that n1 =1.00 for air, and n2 = 1.333 for soap (equivalent to water), n3 = 1.51 for Plexiglass. There is a / 2 shift for ray 1 reflected from the top surface of the bubble, and no shift for ray 2 reflected from the bottom surface. To get constructive interference, then, the path length difference ( 2t ) must be a half-integral multiple of the wavelength—the first three being n / 2, 3n / 2 , and 5n / 2 . To get destructive interference, the path length difference must be an integral multiple of the wavelength—the first three being 0, n , and 2n .

Constructive interference occurs here when

  2tc = n/2 , 3n/2 , 5n/2 , … .

The smallest constructive thickness tc thus is   tc = n/4 = (/n)/4 = (470 nm/1.33)/4 = 88.35 nm

The next thickness that gives constructive interference is tc = 3n/4 , so that    tc' =  265.04 nm

Finally, the third thickness producing constructive interference is tc 5n/4 , so that   tc'' = 441.75

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