- #1

Potatochip911

- 318

- 3

## Homework Statement

The prisms have a refractive index ##n_1=1.3## at wavelength ##\lambda=300nm## and are connected by a material with refractive index ##n_2##. What is the refractive index ##n_2## so that light of wavelength ##>\lambda## follow path 2, and light at ##<\lambda## follow path 3?

## Homework Equations

##\sin\theta_c=\frac{n_2}{n_1}##

##n=\frac{c}{v}=\frac{c}{f\lambda}##

## The Attempt at a Solution

I am very confused as to how the wavelength of light effects total internal reflection. I can solve for the frequency of the light by using ##n=\frac{c}{f\lambda}\Rightarrow f=\frac{c}{n\lambda}## but I'm not entirely sure this is useful. I also have that ##\sin\theta_c=\frac{n_2}{n_1}## which can be rewritten as ##\sin\theta_c=\frac{\lambda_1}{\lambda_2}## The critical angle is a constant in this case I think? it's 45 degrees since the light is always coming in parallel to the bottom of the prisms but I don't understand how varying ##\lambda## is going to change whether or not it exits the prisms.