Q14DQ - University Physics with Modern Physics

Question

 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?

Step-by-Step Solution

Answer

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

Step 1: (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,

$\mathrm{d}=\mathrm{m\lambda }, \mathrm{m}=0,\underline{+1,+2}$                                            ...(i)

 

For destructive interference, the condition for path difference is,

$\mathrm{d}=\left(\mathrm{m}+\frac{1}{2}\right)\lambda , \mathrm{m}=0,\underline{+1,+2}$

                                ...(ii)

Here,  $\mathbf{d}$  is the path difference between two waves and   $\mathbf{\lambda }$  is the wavelength of light used.

Step 2: (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.