How Do You Determine the Angle That Maximizes Light Dispersion?

[chipotleaway]

Homework Statement

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.

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

 

[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.

 

[haruspex]

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.

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

 

[chipotleaway]

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

 

[haruspex]

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

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?

 

[chipotleaway]

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

 

[haruspex]

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

Yes.

 

[chipotleaway]

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 can't be right. i treated each variable as a constant when differentiating with respect to the other

 

[haruspex]

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 can't be right. i treated each variable as a constant when differentiating with respect to the other

<br /> Fixing the LaTex:<br /> \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}<br /> I think you have a sign wrong there, but let&#039;s step back a bit. What do you get for ∂θ₂/∂λ as a function of λ, θ₁ and n(λ)?

 

[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..

 

[haruspex]

I got \frac{sin(\theta_1)}{\sqrt(1-(n(\lambda)sin(\theta_1))^2}\frac{dn(\lambda)}{d\lambda}.

(I corrected your d/dλ term.)
Right, now consider how that will behave as θ₁ increases from 0.

 

[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

 

[UltrafastPED]

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

 

[chipotleaway]

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

 

[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?

 

[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

 

[UltrafastPED]

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

 

[haruspex]

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

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

 

[chipotleaway]

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

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 realized 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.

 

[UltrafastPED]

Max dispersion occurs as you near the critical angle.

 

[chipotleaway]

So the best angle of dispersion is 'infinitely' close to the critical angle..?

 

[UltrafastPED]

Test your hypothesis!

 

[haruspex]

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).

No, don't set n(λ)=1. That's what gave you a result beyond the critical angle. Look at this again:
\frac{sin(\theta_1)}{\sqrt(1-(n(\lambda)sin(\theta_1))^2)}\frac{dn(\theta_2)}{d\theta_1}.
As $\theta_1$ approaches the critical angle, what value does its sine tend to? What does that do to the denominator?

 

[chipotleaway]

is that an error in your derivative term?

letting the incident angle approach the critical angle which is arcsin(\frac{1}{\lambda}[\itex], i get a 0 in the square root and 1/n in the numerator. (from evaluating the limit)

 

[haruspex]

is that an error in your derivative term?

Sorry, yes. The last bit should be dn/dλ, but it isn't important.

letting the incident angle approach the critical angle which is arcsin(\frac{1}{\lambda}[\itex], i get a 0 in the square root and 1/n in the numerator. (from evaluating the limit)

<br /> Exactly. So, as θ₁ increases towards the critical angle, the dispersion increases without limit. The maximum is therefore at the critical angle.

 

[chipotleaway]

Thank you!

 

[haruspex]

Thank you!

You're welcome. Reaching for d/dx = 0 to find an extremum without thinking about absolute extrema is an easy trap to fall into. I went that way at first on this problem too.