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If a light of wavelength $\lambda $ is incident normally and the thickness of film is $t$, then optical path difference between waves reflected from upper and lower surface of the film is :



(A) $2{n_1}t$
(B) $2{n_1}t - \dfrac{\lambda }{2}$
(C) $2{n_1}t + \dfrac{\lambda }{2}$
(D) $2t$

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Last updated date: 20th Jun 2024
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Answer
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HintWhen the light is incident on some surface, some part of the light is reflected by the surface and some part of the light is refracted by the surface. The refracted light is again reflected by the second medium, so the refracted light from the first surface is again reflected from the second surface. Then, the optical path difference is determined by using the optical path formula.

Useful formula
The optical path difference or the optical path length is given by,
$x = n \times t$
Where, $x$ is the optical path difference or optical path length, $n$ is the refractive index of the medium and $t$ is the thickness of the medium.

Complete step by step solution
Given that,
The wavelength of the incident light is, $\lambda $
The thickness of the film is, $t$
The optical path difference or the optical path length is given by,
$x = n \times t\,................\left( 1 \right)$
Here the light is incident to the one surface, so that the light gets reflected and refracted from the upper layer of the surface, and the refracted light is again reflected from the lower surface of the same medium, then the path difference or the path length of the two reflected light rays is given by,
$x = 2{n_1}t$
Where, ${n_1}$ is the refractive index of the first medium.
The value $2$ in the above equation represents that the two times the light is reflected, so the term $2$ is multiplied.

Hence, the option (A) is the correct answer.

Note: When the light is reflected from the two surfaces, then it is known as thin film interference. When the light is reflected from the upper surface and the lower surface of the same film, then we see the different coloured patterns.