homework and exercises - Light of a specific wavelength going through a prism

I won't post the entire question here since I would just like a bit of help getting started as I am quite lost. The question is essentially saying that a light ray parallel to the x axis with a wavelength of $\lambda$ is heading into an isosceles triangle with an index of refraction $n_2$ from air with an index of refraction $n_1$. I'm supposed to find the value of $n_1$ so that light of wavelength $>\lambda$ will exit the prism and light of wavelength $<\lambda$ will be reflected perpendicular to the x axis. I am given the value for $n_2$, $\lambda$ and the angles inside the prism.

The main thing I don't understand is how the wavelength is effecting whether or not the light exits the prism. I believe this depends on whether or not the light is experiencing total internal reflection but in this scenario that would be given by $sin\theta _c=\frac{n_2}{n_1}$ I also know that $n=\frac{c}{v}=\frac{c}{f \lambda }\Rightarrow \sin\theta_c=\frac{\lambda_1}{\lambda_2}$ since only wavelength changes in the different mediums. So the the critical angle is clearly affected by the wavelength of light in the different mediums but I just can't figure out how this relates to higher or lower frequencies exiting or staying in the prism.

Answer

To complement Samuel Weir's answer, look at this image of incandescent (white) light passing through a prism.

As can be seen violet light is refracted ('bent') more than red light. Violet light is more energetic than red light and we know that:

$\lambda=\frac{hc}{E}$. ($h$ is Planck's constant, $c$ the speed of light, $E$ is energy).

So violet light is of shorter wavelengths (larger $E$). Light of shorter wavelength is thus more refracted by a prism than light of longer wavelength.

Edit:

Following the asker's comments I understand his problem to be as in the following diagram:

For rays following path I, we get refraction:

$n_2\sin\theta_1=n_1\sin\theta_2$

Or: $\sin\theta_1=\frac{n_1}{n_2}\sin\theta_2$

For rays following path II we have total reflection:

Acc. Snell's Law, for total reflection, $\theta_2=90^0$, so $\sin\theta_2=1$.

So $\sin\theta_1=\frac{n_1}{n_2}$ and $n_1=n_2\sin\theta_1$.

Because the triangle is isosceles, $\theta_1=45^0$, so:

$\large{n_1=\frac{\sqrt{2}}{2}n_2}$

This the critical value for $n_1$, in reality to get total refection the condition:

$\large{n_1<\frac{\sqrt{2}}{2}n_2}$ must be met.