Abeam of light passing through a flint glass is incident on water. the index of refraction of water is 1.33 and the index of refraction of flint glass is 1.62. a. what is the critical angle of light that passes from the flint glass to water? b. what happens to the path of light if the angle of incidence is 60º? c. what happens to the value of the critical angle if the flint glass is replaced by a diamond? the index of refraction of diamond is 2.42.

(a) when light travels from a medium with high refractive index to a medium with lower refractive index, there exist a critical angle above which there is no reftaction, but all the light is reflected, and the value of this angle is given by:

where n2 is the refractive index of the second medium, while n1 is the refractive index of the first medium.

In this part of the problem, the light travels from flint glass () to water ), so the value of the critical angle is


(b) We can solve the problem by using Snell's law:

where
is the reafractive index of the first medium
is the angle of incidence
is the refractive index of the second medium
is the angle of refraction

In this part of the problem, the light travels from flint glass () to water ), and the angle of incidence is , so we can re-arrange the equation to calculate the angle of refraction:


However, there is no solution for this equation. In fact, the angle of incidence is above the critical angle we found at point a) (): this means that there is no refraction, so the light is not transmitted into the water, but it is only reflected.

c) If we replace the refractive index of flint glass (1.62) with the refractive index of diamond (2.42), the new critical angle is

so, the angle of incidence has decreased.

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