# Two Prisms connected by a material of refractive index n

1. Sep 28, 2015

### Potatochip911

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
$\sin\theta_c=\frac{n_2}{n_1}$
$n=\frac{c}{v}=\frac{c}{f\lambda}$

3. 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.

2. Sep 28, 2015

### Daz

The question only says that n1 is 1.3 at a particular wavelength: 300nm. Do you suppose it could be different at other wavelengths? How would it vary?

3. Sep 28, 2015

### Staff: Mentor

The refractive index depends on the wavelength, so the critical angle depends on it as well. Here, the transition between total reflection and transmission is supposed to be at a wavelength of 300 nm, so you can calculate the necessary refractive index at 300 nm to be directly at the transition.

4. Sep 28, 2015

### Potatochip911

So the correct refractive index then is $n_1=\frac{n_2}{\sin\theta_1}$, edit: I mean $n_2=n_1\sin\theta_1$ sorry for all the mistakes I'm on my phone