Critical angle of refraction

[sgoeke]

Homework Statement

A ray of light travels across a glass-to-liqued interface. If the indices of refraction for the liquid and glass are, respictevly 1.75 and 1.52, what is the critical angle at this interface?

Homework Equations

critical angle = inverse sin (n2/n1)

The Attempt at a Solution

i used inverse sin (1.52/1.75) and got an angle of 60.3 degrees. However, I do know that if you reverse the 1.75 and 1.52 in that equation, you get an undefined angle and the correct answer is that there is not critical angle. Why do you used 1.75/1.52?

 

[Stonebridge]

Homework Statement

A ray of light travels across a glass-to-liqued interface. If the indices of refraction for the liquid and glass are, respictevly 1.75 and 1.52, what is the critical angle at this interface?

Homework Equations

critical angle = inverse sin (n2/n1)

The Attempt at a Solution

i used inverse sin (1.52/1.75) and got an angle of 60.3 degrees. However, I do know that if you reverse the 1.75 and 1.52 in that equation, you get an undefined angle and the correct answer is that there is not critical angle. Why do you used 1.75/1.52?

You get a critical angle and total internal reflection when the light goes from a dense to a less dense medium. That is, one with a refractive index that is higher to one that is lower.
The question says that the light goes from glass to liquid and that glass has index=1.52 and the liquid 1.75.
So there is no critical angle in this case.

The law is n₁ sin i₁ = n₂ sin i₂

For the critical angle, i₂ = 90 so, sin i₂ = 1