Question

1) Calculate the refractive index of water if a light ray incident on a glass-
water interface at an angle of 45degree suffers a deviation of 15degree.
(Refractive Index of glass=1.4)​

Answer 1

Answer:

45.2 degrees

Because

Given that,

Incident angle i=45 degrees

Let critical angle be ic

Now,

sin ic =μ

Where, μ is the refractive index of glass

Now, again

r=45 degrees − 15 degrees

r=30 degrees

sin r / sin i =μ

μ= sin 45 degrees/ sin 30 degrees

μ= \sqrt{2}

2

μ=1.4

Now, the critical angle is

sin ic = 1/μ

sin ic = 1/1.4

ic = sin^{-1}sin

−1

(0.71)

ic =45.2 degrees

Hence, the critical angle is 45.2 degrees

Explanation:

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