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A very thin soap film (n = 1.33), whose thickness is much less than a wavelength of visible light, looks black; it appears to reflect no light at all. Why? By contrast, an equally thin layer of soapy water (n = 1.33) on glass (n = 1.50) appears quite shiny. Why is there a difference?

Short Answer

Expert verified

In the first case, the interference is destructive but in the second case, the interference is constructive.

Step by step solution

01

(a) Concept of interference.

It is a phenomenon where one light wave superposes over another coherent light wave and this superposition leads to a redistribution of the intensity of light rays at different points.

For constructive interference, the condition is,

d=³¾Î»,m=0,+1,+2 ...(i)

For destructive interference, the condition for path difference is,

d=m+12λ,m=0,+1,+2

...(ii)

Here, dis the path difference between two waves and λ is the wavelength of light used.

02

(b) Explanation of the argument.

For very thin soap film, the beam gets reflected first at the air-water boundary and the second reflection comes from the water-air interface. Both of these beams that are reflected from different interfaces have the same phase and cause destructive interference. Therefore, the film has a dark appearance.

For thin soap film on glass, the reflection happens at three interfaces- air-water, water-glass and glass-air interface. The third reflection causes the overall interference to be constructive and thus the film appears to be shining.

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