Mumbai university > FE > SEM 2 > Applied Physics II

Marks: 7M

Year: May 2015

• 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.

$$Ґ = 2µd$$

• At oblique incidence the path difference is given by

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

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

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

• 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

$$2µd cos r + \frac{λ}{2} = nλ$$

$$2µd cos r = nλ - \frac{λ}{2}$$

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

• The film will appear dark in the reflected light when

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

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