Calculate the best angle for maximum light dispersion through a medium

  • #1
Jan Berkhout
8
3
Homework Statement:
If you have light moving from ##n_1## to ##n_2## you can get dispersion if ##n_1## is a function of wavelength.
What angle of incidence ##(θ_1)## will maximise the dispersion for the situation below where the light goes from a medium with ##n_1(λ)## to vacuum, ##(n_2 = 1)##?
Relevant Equations:
$$\frac{d \sin ^{-1}(\text{ax})}{\text{dx}}=\frac{a}{\sqrt{1-(ax)^2}}$$ $$\frac {dy}{dx}=\frac{dy}{dz} \frac{dz}{dx}$$
I first thought that the angle would have to be maximum when it is closest to the critical angle for total internal reflection. From my lectures the equation for the critical angle is ##\theta _1>\ sin ^{-1} \left( \frac {n_2} {n_1} \right),## so as ##n_2 = 1##, we have ##\theta _1=\sin ^{-1}\left(\frac{1}{n_1(\lambda)}\right)##. I didn't really know what to do after that but from the equations given in the hint (relevant equations), I thought I'd have to differentiate with respect to ##\lambda##. This gives $$\frac {d θ_1} {d \lambda} = -\frac{\frac{d}{{d \lambda}} n_1(\lambda)}{n_1(\lambda) \sqrt{n_1(\lambda)^2-1}}.$$ So I have the change in the critical angle with respect to the change in wavelength, so my hunch is I have to set the derivative to 0 and solve for \lambda to find the wavelength for the maximum angle then I can calculate the angle? Is this right?
 
Last edited:

Answers and Replies

  • #2
Jan Berkhout
8
3
I worked hard and actually believe I have solved it! So this thread can be closed :)
 
  • #3
haruspex
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This gives $$\frac {d θ_1} {d \lambda} = -\frac{\frac{d}{{d \lambda}} n_1(\lambda)}{n_1(\lambda) \sqrt{n_1(\lambda)^2-1}}.$$
Don't you want to maximise ##\lambda## wrt ##\theta_1##? For that you need $$\frac {d \lambda} {d θ_1}$$.
 

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