Question

Calculate the critical angle of glass-water surface whose refractive index is 1.5 and 1.33

Answer 1

Explanation:

Given that refrective index=4/3

Let critical angle=C

According to Snell's law

C=sin^-1(1/refrective index)

C=sin^-1(3/4)

C=sin^-1(0.75)

C=48.59°

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Tipper Rumpf, Ph.D. Optics & Computational Physics, University of Central Florida (2006)

Answered April 11, 2018 · Upvoted by Satya Parkash Sud, M.Sc. Physics & Nuclear Physics, University of Delhi (1962)

Originally Answered: The refractive index of water is 4/3. How would you calculate the critical angle of the water-air interface?

Refraction at an interface is described by Snell’s law.

n1sinθ1=n2sinθ2n1sin⁡θ1=n2sin⁡θ2

If n1n1 is the higher refractive index medium, we will have a critical angle. If θ1θ1 is set to be exactly at the critical angle θcθc, then θ2θ2 will be exactly at 90∘∘. Substituting all of this into Snell’s law give us

n1sinθc=n2sin90∘n1sin⁡θc=n2sin⁡90∘

Solving this for θcθc yields

θc=sin−1(n2/n1)θc=sin−1⁡(n2/n1)

In this case, n1=1.33n1=1.33 because it is water and n2