# 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.

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.

• Okay I am slightly confused about something else now. So the question is worded as "you have a prism with $n_2$ at the wavelength $\lambda$". And from your post and Samuel's post it sounds to me as if $n_2$ and $\lambda$ are both variable depending on whats going on. So what I'm curious about is does this mean that the frequency $\lambda$ is for when it's outside the prism or inside the prism. I feel like I'm wording this very poorly but hopefully you can see what I'm confused about. – Jeffrey Sep 27 '15 at 0:16
• @Jeffrey: the speed of light is lower in optically dense materials (like glass) than in vacuum (or air), so the angle of refraction depends on the refractive indices of both media (glass and air here). The RI for glass does depend slightly on wavelength (in vacuum), which is what causes the dispersion (colour separation). Might this link help physicsclassroom.com/class/refrn/Lesson-4/… ? – Gert Sep 27 '15 at 0:47
• I am currently stuck with this: $n_1\sin\theta_1=n_2\sin\theta_2$, I want to find the value at which TIR occurs so $\sin\theta_2=1$ then I get $\sin\theta_1=\frac{n_2}{n_1}$, but in this question $\theta_1$ is a constant that doesn't change. So is it reasonable that I can find $n_1$ by using $n_1=\frac{n_2}{\sin\theta_1}$? The problem with this is that it's giving me quite an odd answer with n_1>n_2 so I don't think it's correct. – Jeffrey Sep 27 '15 at 2:12
• @Jeffrey: I think you need to expand your question a bit. W/o a diagram it's really hard to see what you're trying to solve. I'll bookmark this question and revisit tomorrow. It's very late here now. – Gert Sep 27 '15 at 2:19
• I drew a picture of it here to show what I mean by the angle $\theta_1$ being a constant. Diagram: so as you can see from the image I have to determine $n_1$ so that light of wavelength larger than $\lambda$ will exit along path 2 and wavelength smaller than $\lambda$ will be reflected about path 3. Path #2 looks parallel to the original but that's just me being a poor artist. – Jeffrey Sep 27 '15 at 7:53

The main thing I don't understand is how the wavelength is effecting whether or not the light exits the prism.

Wavelength comes into play because the index of refraction of a material is not a fixed constant. It depends on the wavelength. That's why a prism of glass breaks white light into a rainbow of colors. The different wavelengths of light that make up what we see as 'white light' are "bent" (or "refracted") by different amounts by the glass prism because the index of refraction of the glass is not a constant n but rather a wavelength dependent n(λ).

I believe that the index of refraction of most glasses in the visible wavelength region tends to increase with wavelength (or is it the other way around? I'll leave that to you to look up). So if you find the key wavelength which is at the borderline between allowing light to either (1) exit the prism or (2) "reflect perpendicular to the x axis", then it follows that wavelengths shorter and longer than this key wavelength will behave as (1) and (2), respectively; or as (2) and (1), respectively.