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The reflection of perpendicularly incident white light by a soap film in the air has an interference maximum at 600nmand a minimum at role="math" localid="1663024492960" 450nm, with no minimum in between. If n=1.33for the film, what is the film thickness, assumed uniform?

Short Answer

Expert verified

The thickness of the soap film is 338nm.

Step by step solution

01

Given Data.

  • The refractive index of the thin film isn2=1.33
  • The maximum interference is observed atlocalid="1663026501026" role="math" λmax=600nm
  • The minimum interference is observed atlocalid="1663026510207" λmin=450nm
02

Interference in thin films.

Bright colours reflected from thin oil on water and soap bubbles are a consequence of light interference. Due to the constructive interference of light reflected from the front and back surfaces of the thin film, these bright colours can be seen. For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:

2L=m+12λn2m=0,12..(Maxima—bright film in the air)

where λis the wavelength of the light in air, Lis its thickness, and n2is the film’s refractive index.

Here, in this case, the air is on the both sides of the soap film, therefore the conditions for constructive and destructive interference.

2L=mλminn2 (Destructive)

2L=m+12λmaxn2 (Constructive)

03

Determining thickness for the first few orders.

For constructive interference, the value of thickness for the first few orders is

m=0;L=0+12600nm21.33=113nmm=1;L=1+12600nm21.33=338nmm=2;L=2+12600nm21.33=564nm

For destructive interference, the value of thickness for the first few orders is

m=1;L=1450nm21.33=169nmm=2;L=2450nm21.33=338nmm=3;L=3450nm21.33=254nm

.

The least thickness, which is common for constructive and destructive interference, is 338nm. Hence the thickness of the soap film is 338nm.

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