Light of wavelength 584 nm is incident normally on a film of water 1.0 µm thick.

Question

Light of wavelength 584 nm is incident normally on a film of water 1.0 µm thick. The index of refraction of water is 1.33.

(a) What is the wavelength of the light in the water (nm)?

(b) How many wavelengths are contained in the distance 2t, where t is the thickness of the film? (Do not round your answer to a whole number.)

(c) What is the phase difference between the wave reflected from the top of the air-water interface and the one reflected from the bottom of the water-air interface in the region where the two reflected waves superpose (in radians)?

Explanation / Answer

a) wavelength is compressed in media. Actually, we usually specify the 584nm as vacuum wavelength or air wavelength too, but air and vacuum are close enough to equal that it doesn't matter here. 584nm (1/1.33) = 439.0977444 nm or 439 nm = 439 E-9 m b) assume we are going straight into film, so 2t = 2 µm = 2 E-6 m 2µm/439 nm = 4.554794521 wavelengths c) going from low wavelength to high wavelength there is a 180 deg (or Pi radians)phase shift when light is reflected, so the light that reflects off the top surface has a 180 deg phase shift. The light that is transmitted into the film and then reflects off the bottom water-air interface does not have this phase shift. So the phase difference will be (4.554794521-4)*360 - 180 = 19.73 deg The reason we subtracted 4 from the number of wavelengths is that every whole number of wavelengths resets the phase. So only the fractional wavelengths will give a measureable phase shift.

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