The total internal reflection between 2 materials is seen to occur if a critical angle of 60 deg. is exceeded.

Question

If the index of refraction of the material that light is entering is 1.98, find the index of refraction of the material in which the light is originating.

Answer 1

If a ray passing through the interface makes angles of Θ1 and Θ2 (with respect to the normal) in materials with refractive indices of n1 and n2 then ,by Snells Law,
sin(Θ1)/n1 = sin(Θ2)/n2. So

sin(60)/n1 = 1/1.98 and n1 = 1.98(1/2)√3 = 1.715


Doug

Answer 2

If I'm reading your question correctly, Doug (the previous answerer) has it backwards. My interpretation of what you are asking is: "If a light ray is traversing medium 1, with refractive index n1, and encounters an interface with medium 2 that has refractive index of n2 = 1.98, the critical angle is found to be 60 degrees. What is the unknown refractive index 'n1'?"

We know that n1 > n2, because that is the only situation for which total internal reflection can occur, so the previous answer must be incorrect (he had n1 < n2).

For the situation rescribed above, the critical angle is given by (see source):

sin(theta_critical) = n2/n1; n1 > n2

Here, n2 = 1.98, and theta_critical = 60 degrees.


Rearranging the above equation to solve for n1 yields:

n1 = n2/sin(theta_critical) = 1.98/sin(60 degrees)

n1 = 1.98/(sqrt(3)/2) = 2.286