Derive the condition for a thin transparent film of constant thickness to appear bright and dark when viewed in reflected light.

48views

written 8.6 years ago by

teamques10 ★ 69k

Let us consider a plane transparent film as shown in the figure below.
Let light be incident at A.
Part of the light is reflected towards B and other part is refracted into the film towards C.
This second part is reflected at C and emerges at D and is parallel to the first part.
At normal incident, the path difference between rays 1 and 2 is twice the optical thickness of the film.

$Ґ = 2µd$

$Ґ = µ (AC + CD) - AB = \frac{2µd}{cos⁡r} - AB$

$Ґ = \frac{2µd}{cos⁡r} - 2µd tan r sin r$

$Ґ= 2µd (\frac{1}{cos⁡r} - tan r sin r) = 2µd \frac{1- sin^2 r}{cos⁡r} = 2µd cos r$

Where µ is the refractive index of the medium between the surfaces.
Since for air µ=1, the path difference between rays 1 and 2 is given by
$Ґ=2d cos r$
While calculating the path difference, the phase change that might occur during reflection has to be taken into account.
Whenever light is reflected from an interface beyond which the medium has lower index of refraction, the reflected wave undergoes no phase change.
When the medium beyond the interface has a higher refractive index there is phase change of π.
The transmitted waves do not experience any phase change.
Hence, the condition for maxima for the air film to appear bright is

$2µd cos r + \frac{λ}{2} = nλ$

$2µd cos r = nλ - \frac{λ}{2}$

$2µd cos r = (2n-1) \frac{λ}{2}.....where \ \ n = 1, 2, 3,…$

$2µd cos r + \frac{λ}{2} = (2n-1)\frac{λ}{2}$

$2µd cos r = nλ......where \ \ n=1, 2, 3,…$

ADD COMMENT EDIT