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A parallel beam of light of wavelength 560 nm is incident on a thin oil film (refractive index 1.4) surrounded by air. What is the minimum film thickness for which the reflected light is weak (i.e., the transmitted light is strong / weakly absorbed in reflection)? Equivalently, find the minimum thickness for which the film gives minimum reflection (maximum transmission).

1. 100 nm
2. 200 nm
3. 300 nm
4. 400 nm

Correct answer: 200 nm

Solution

Light reflecting from the top air-oil surface undergoes a pi phase change; the bottom oil-air reflection does not. For destructive interference in reflection (weak reflection, strong transmission), the condition is 2*mu*t = m*lambda. The smallest nonzero thickness uses m = 1: t = lambda/(2*mu) = 560/(2*1.4) = 560/2.8 = 200 nm.

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