Question

A ray of light in air is incident on an air-to-glass boundary at an angle of 30 degrees with the normal if the index of refraction of the glass i 165
what is the angle of the refracted ray within the glass with respect to the normal?​

Answer 1

Explanation:

Use Snell's law:

Use Snell's law:ni * sine(Theta i) = nr * sine(Theta r)

Use Snell's law:ni * sine(Theta i) = nr * sine(Theta r)where

Use Snell's law:ni * sine(Theta i) = nr * sine(Theta r)whereni =1.00 (in air), Theta i=30. degrees, nr =1.65

Use Snell's law:ni * sine(Theta i) = nr * sine(Theta r)whereni =1.00 (in air), Theta i=30. degrees, nr =1.65Substitute and solve for Theta r.

Use Snell's law:ni * sine(Theta i) = nr * sine(Theta r)whereni =1.00 (in air), Theta i=30. degrees, nr =1.65Substitute and solve for Theta r.sine(Theta r) = 1.00 * sine(30. degrees) / 1.65 = 0.3030

Use Snell's law:ni * sine(Theta i) = nr * sine(Theta r)whereni =1.00 (in air), Theta i=30. degrees, nr =1.65Substitute and solve for Theta r.sine(Theta r) = 1.00 * sine(30. degrees) / 1.65 = 0.3030Theta r = invsin(0.3030) = 17.6 degrees