Michaelson Interferometer and changing length of glass rod

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Homework Statement


The index of refraction of a glass rod is 1.48 at T=20°C and varies linearly with temperature, with a coefficient of 2.5 E -5 °C-1. The coefficient of linear expansion of the glass is 5 E -6 °C-1. At 20.0°C the length of the rod is 3.00 cm. A Michelson interferometer has this glass rod in one arm, and the rod is being heated so that its temperature increases at a rate of 5.00 °C/min. The light source has wavelength λ = 599nm, and the rod initially is at 20°C.

How many fringes cross the field of view each minute?

Homework Equations



y = [itex]\frac{mλ}{2}[/itex]
Where y is the change in distance.
m is the number of fringes.
λ is the wavelength
(just for simplicity)
α = 5 E -6 °C-1
β = 2.5 E -5 °C-1

The Attempt at a Solution


I've attempted this problem several times. I had gotten it correct once before, but can't seem to do it this time (practicing it for my upcoming final).
First, I find the change in distance. 5α since the temperture rises by 5°C in one minute.
Now, the change in the index of refraction: 5β.
Since the equation deals with the change in distance rather than the total distance, we plug in to the original equation.

(5α) = .5m(λ/(n*5β) or, solving for m
m = [itex]\frac{2*5α cm*1.48*5β}{599nm}[/itex]
Make the units equal, and the answer is wrong.

So I repeated my steps, this time setting the temperture to 25°C. No dice.
Tried multiplying the length to 5α and both methods (°C = 5 and 25). But Nope.avi

The answer should be around 14 I believe. Some guidance and explanation would be much appreciated.
 

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