Question

A light ray is incident on air-liquid interface at 45° and is refracted at 30°.what is the refractive index of the liquid? For what angel of incidence will the angle between reflected ray and refracted ray be 90°?

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Answer 1

Hello dear!!!..

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Answer 2

Thanks for asking the question!

ANSWER::

Given:-

Angle of incidence , ∠i = 45°

Angle of refraction , ∠r = 30°

Now , Refractive Index (n)

n = sin i / sin r

n = sin 45° / sin 30°

n = (1/√2) / (1/2)

n = √2

Given => Angle between reflected and refracted ray is 90°

So ,

Angle of reflection (∠R) + Angle of incidence (∠i) = 90°

Angle of refraction (∠r) = 90° - ∠i

And , refractive index (n) = sin i / sin r

n = sin i/ sin(90-∠i)

n = sin i / cos i [sin(90 -Ф) = cos Ф]

Now , as we have got √2 above lets see ,

tan i = √2

tan i = 1.414

And if we look from the trigonometric ratios table we will get ,

∠i = 54.7° (approximate value)

Hope it helps!