Finding the Critical Angle: Diamond and Water

[AlanxD]

Homework Statement

The critical angle diamond (n=2.42) and water (n=1.33). Find the angle.

Homework Equations

The Attempt at a Solution

I dun get how u find the angle with 2 n's. I need a solution.

 

[radou]

Perhaps I'm missing something, but isn't there one single expression for the critical angle, dependent only upon the two indices of refraction?

 

[AlanxD]

Well the question is 2 sentences. Critical angle of diamond is n=2.42. Water n=1.33. Find the angle of incidence. My textbook says the angle is 33 degrees but i do not knw how it arrived at that answer.

 

[radou]

I hope http://www.glenbrook.k12.il.us/GBSSCI/PHYS/Class/refrn/u14l3c.html" helps.

 

[AlanxD]

Ya i knw u can find the critical angle but i got the critical angle of diamond and n for water. I want to knw how to solve the angle between diamond and water.

 

[andrevdh]

Snell's law still applies to the situation.

The incident angle is in the diamond and the refraction angle is in the water.

At the critical incident angle (in the diamond) the refraction angle (in water) is equal to ninety degrees.

From this info you can set up an equation using Snell's law (at the diamond-water interface) and solve for the critical incident angle - just show try and we will help you to set it up.

 

[AlanxD]

I found snails law. NiSinangle of Incidence=Nrsinangle of refraction. What should i do now?

 

[andrevdh]

Snell's law is applied at the interface where the refraction takes place. In this case light is refracted from a diamond into water. For total internal refraction to be able take place the light must travel from a more to a less optical dense medium, which is what we have here since the refractive index of diamond is larger than that of air.

In this case the incident refractive index is in diamond so n_i is that of diamond and n_ris that of water. When the incident angle of the light in diamond is equal to the critical incident angle \theta _c we find that the refractive beam of light is at the point of dissappearing - it is traveling along the interface. This means that the angle of refraction (in air) will be ninety degrees when the incident angle is equal to the critical incident angle (if the incident angle is made larger than the critical angle we find the refracted beam dissappears and we have total internal reflection). Fill this information into Snell's equation and see if you can solve for the critical incident angle - if not just show Snell's law with the values plugged in.