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Homework Statement
Show that if cosΦ is replaced by its third-degree Taylor polynomial in Equation 2, then Equation 1 becomes Equation 4 for third-order optics. [Hint: Use the first two terms in the binomial series for [itex]ℓ^{-1}_o[/itex] and [itex]ℓ^{-1}_i[/itex]. Also, use Φ ≈ sinΦ.]
Homework Equations
Sorry that these don't look very nice. I'm pretty sure all the parentheses are necessary. I've been trying to learn as I go so I didn't know all the commands to make it look nicer. It's taken me well over 30 minutes to type this whole thread out (I was experimenting so yeah...). I'd recommend just looking at the screenshots from my pdf.
Equation 1:
[itex]n_1/ℓ_o + n_2/ℓ_i = (1/R)(n_2s_i/ℓ_i - n_1s_o/ℓ_o)[/itex]
Equation 2 (comes in a pair):
[itex]ℓ_o = √(R^2 + (s_o + R)^2 - 2R(s_o + R)cosΦ)[/itex]
[itex]ℓ_i = √(R^2 + (s_i - R)^2 + 2R(s_i - R)cosΦ)[/itex]
Equation 4:
[itex]n_1/s_o + n_2/s_i = (n_2 - n_1)/R + h^2( (n_1/(2s_o))(1/s_o + 1/R)^2 + (n_2/(2s_i))(1/R - 1/s_i)^2)[/itex]
The Attempt at a Solution
Third-degree Taylor polynomial for cosΦ is:
[itex]1 - Φ^2/2![/itex]
I plugged in Φ ≈ sinΦ into the third-degree polynomial:
[itex]1 - sin^2Φ/2[/itex]
Plugged that into Equation 2 and got:
[itex]√(R^2sin^2Φ + Rs_osin^2Φ + s^2_o)[/itex]
[itex]√(R^2sin^2Φ - Rs_isin^2Φ + s^2_i)[/itex]
Then I plugged that into Equation 1. I don't know what to do after this. I know the hint says to use the first two terms in the binomial series for [itex]ℓ^{-1}_o[/itex] and [itex]ℓ^{-1}_i[/itex], but I have no clue how I would do that since it's not in the form where I could use the binomial series. I have tried manipulating it so it would be, but nothing has worked.
Side Note:
This is the first time I have posted on this forum so I apologize if I have broken any rules.