Deducing the film thickness of this ques below:

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    Film Thickness
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The discussion focuses on deducing the film thickness of a thin soap film observed in sodium light, which appears dark due to interference effects. The darkness results from light waves reflecting off the front and back faces of the film, canceling each other out when there is a path difference of λ/2. A phase change occurs for light reflecting off the air/film surface, while light from the film/air surface does not experience this change. Therefore, a very thin film, less than one wavelength, will not reflect light, indicating its presence just before bursting. The next thickness that results in darkness is λ/2, as the total path difference accounts for the phase shift at the front surface.
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a thin soap film is seen in sodium light by normal reflection method it appears totally dark deduce the possible values of film thickness given (myu)-refra.indx of liquid-1.333 and lambda wave.lgth-5893 angstrom
 
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This effect is caused by INTERFERENCE between light waves reflected from the front face of the soap film and light waves reflected from the back face, after passing into the film.
The waves will 'cancel out' if there is a path difference of λ/2 (a phase difference of ∏)
and the film will look dark because there is no light reflected.
There is a strange fact about waves. Light reflected from the air/film surface undergoes a phase change of ∏, this is like being given a λ/2 kick out of step.
Light reflected from the film/air surface does not experience a phase change.
This means that a very thin film (less than 1 wavelength) will produce no reflected light.
This is why you can be aware of the dark film when the film is about to burst.
The next thickness of film that will produce darkness is when the film is λ/2 thick (this is the λ in the film material) because the light travels 1 wavelength through the film (there and back)
and the resulting total path difference is λ/2 because of the λ/2 at the front surface.
 
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