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A ray is incident on a film of thickness t and the index of refraction n=1.6 at an angle θ = 35 degrees. Find the angle of refraction in the film: ________degrees. If the reflected ray comes out at a distance___________ 2.5 mm. What is the thickness of the film?

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Answer

Given,

refractive index of film, n = 1.6

refractive index of air, n' = 1

angle of incidence, i = 35°

angle of refraction, r = ?

Using Snell's law

n' sin i = n sin r

1 x sin 35° = 1.6 x sin  r

 r = 21°

Angle of refraction is equal to 21°.

Now,

distance at which refractive angle comes out

d = 2.5 mm

α be the angle with horizontal surface and incident ray.

α = 90°-21° = 69°

t be the thickness of the film.

So,

[tex]tan \alpha = \dfrac{t}{d/2}[/tex]

[tex]tan 61^0 = \dfrac{t}{2.5/2}[/tex]

t = 2.26 mm

Hence, the thickness of the film is equal to 2.26 mm.

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