# Homework Help: Maximising light dispersion

1. Nov 1, 2013

### chipotleaway

1. The problem statement, all variables and given/known data
If light moves from a medium with a refractive index that is a function of wavelength, [\itex]n_1(\lambda)[/itex], to vacuum $n_2=1$, then dispersion will occur. Find the incident angle $\theta_1$ that will maximize dispersion.

3. The attempt at a solution
I'm interpreting 'maximum dispersion' as if we change the wavelength a little, then the angle of refraction $\theta_2$ should change a lot. So I think we trying to find the maximum of the function $\frac{d\theta_2}{d\lambda}$, which means solving $\frac{d^2\theta_2}{d\lambda^2}=0$ for $\theta_1$.

Is the correction interpretation and approach? Thanks

2. Nov 1, 2013

### UltrafastPED

The question asks for the _angle_ which results in the most dispersion; your wavelength dependent index of refraction is the dispersion relationship.

So consider the total fan of light for a particular set of wavelengths - for example, the two wavelengths 400 nm and 700 nm. The rays are parallel as they approach the interface, but are "dispersed" by some angle when they leave. It is this "dispersion" angle which needs to be maximized.

You already know the angle of incidence which minimizes it: zero degrees.

3. Nov 1, 2013

### haruspex

You are not trying to maximise $\frac{d\theta_2}{d\lambda}$ wrt λ.

4. Nov 1, 2013

### chipotleaway

So I should be maximizing the difference between the diffraction angles for each wavelengths?

5. Nov 1, 2013

### haruspex

No. You are trying to maximise $\frac{d\theta_2}{d\lambda}$, but not wrt λ. What is the variable for which you are trying to pick a value?

6. Nov 1, 2013

### chipotleaway

$\theta_1$, which would mean $\frac{d}{d\theta_1}\frac{d\theta_2}{d\lambda}$?

7. Nov 2, 2013

Yes.

8. Nov 2, 2013

the function i get is $$\frac{cos(\theta_1)(1-n^2sin^2(\theta_1))-n^2cos(\theta_1)sin^2(\theta_1)}{(1-n^2sin^2(\theta_1))^{3/2)}[\tex]. the numerator simplifies to cos(\theta) which implies the angle is 90 degrees which cant be right. i treated each variable as a constant when differentiating with respect to the other 9. Nov 2, 2013 ### haruspex Fixing the LaTex: [tex]\frac{cos(\theta_1)(1-n^2\sin^2(\theta_1))-n^2\cos(\theta_1)sin^2(\theta_1)}{(1-n^2\sin^2(\theta_1))^\frac 32}$$
I think you have a sign wrong there, but let's step back a bit. What do you get for ∂θ2/∂λ as a function of λ, θ1 and n(λ)?

10. Nov 2, 2013

### chipotleaway

Differentiating $$arcsin(n(/lambda)sin(\theta_1)$$, I got $$\frac{sin(\theta_1)}{\sqrt(1-(n(\lambda)sin(\theta_1))^2}\frac{dn(\theta_2)}{d\lambda}$$.

ah, i seem to have forgotten the derivative term..

Last edited by a moderator: Nov 2, 2013
11. Nov 2, 2013

### haruspex

Right, now consider how that will behave as θ1 increases from 0.

12. Nov 2, 2013

### chipotleaway

Sorry about the latex errors (i am posting from my phone not good with the small keyboard).

It increases, and does so at faster rate for the larger n is...just looking at the plot of the expression without dn/dλ, there's no upper bound

13. Nov 2, 2013

### UltrafastPED

Now compare with Snell's law ... is there a maximum angle of incidence?

14. Nov 2, 2013

### chipotleaway

could you please elaborate? do you mean find the max. angle of incidence from the expression?

15. Nov 2, 2013

### UltrafastPED

Yes; start with Snell's law and find the maximum angle of incidence for a ray which exits the media ... do this for your shortest and longest wavelengths.

Then compare this result with the result of your calculations. Should these give the same results?

16. Nov 2, 2013

### chipotleaway

oh so the max. angle at which the ray exits. Wouldnt that just be maximizing the function of the refracted angle against the incident angle?

there are no numbers given for wavelength so i'm not sure if we're to assume visible light or not

17. Nov 2, 2013

### UltrafastPED

Then just try a couple - 400 nm, 700 nm.

18. Nov 2, 2013

### haruspex

Right, so what incident angle maximises it? (This is not a local maximum. The slope is never 0.)

19. Nov 3, 2013

### chipotleaway

Taking n(λ)=1, the plot suggests it would be as larger an incident angle (x) in the plot as possible, which would be parallel to the boundary. But that's beyond the critical angle (and light along the boundary wouldn't be refracted or reflected, I don't think).

I also realised when trying to plot for larger n(λ) values, as it should be, that the incident angle must decrease as n(λ) does ot else you get negative square roots.
This doesn't make sense to me cause there shouldn't be restrictions on the incident angle.

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20. Nov 3, 2013

### UltrafastPED

Max dispersion occurs as you near the critical angle.