Michelson Interferometer with expanding glass rod
So I've been working on this question for a while and am still no further at finding an answer. I'm probably just approaching this wrong, I'm at a loss for ideas though.
"The index of refraction of a glass rod is 1.48 at T = 20.0°C and varies linearly with temperature, with a coefficient of 2.50 × 10-5/°C. The coefficient of linear expansion of the glass is 5.22 × 10-6/°C. At 20.0°C the length of the glass is 3.00cm. A <PERSON> 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. 797
The light source has a wavelength λ=589nm, and the rod initially is at T = 20.0°C. How many fringes cross the field of view each minute?"
So our values $$ n = 1.48 , T = 20^{\circ} C , L = 0.03m ,\lambda_{0} = 589nm , $$
$$ \beta = 2.510^{-5} / ^{\circ} C , \alpha = 5.2210^{-6} \frac{m}{ ^{\circ} C} , \Delta T = 5^{\circ} \frac{^{\circ} C}{min} $$
So my attempt: I first set about find the change per minute in $ L $,$ n$ and $ \lambda $ .
$$ \Delta L = \alpha \Delta T=(5.22*10^{-6} \frac{m}{ ^{\circ} C})(5^{\circ}C)=5 \alpha $$
$$ \Delta n = \beta \Delta T=(2.5*10^{-5} / ^{\circ} C)(5^{\circ}C)=5 \beta $$
$$ \Delta \lambda = \frac{\lambda_{0}}{\Delta n}= \frac{\lambda_{0}}{5 \beta} $$
I also wrote them as a function if it makes a difference.
$$ T(t)=T_{0} + \Delta T*t $$
$$ n(t)=n_{0} + \Delta n*t $$
$$ L(t)=L_{0} + \Delta L*t $$
$$ \lambda (t) = \frac{\lambda}{ \Delta n*t} $$
So I want to find $ \Delta N $ the number of fringes per minute. So I first tried using $ \Delta L = \frac{N \lambda}{2} $ which came to $ m= \frac{2 \Delta L}{ \Delta \lambda} $ but that didn't work out, the units cancel each other out. Next I tried using $ N \lambda = 2(n-1)L $ solving for $ N $ we get
$$ N = \frac{2L(n-1)}{ \lambda} $$
then from that
$$ \Delta N = \frac{2 \Delta L( \Delta n-1)}{ \Delta \lambda} $$ which comes to $ \Delta N = -0.001108 /min $ .
The units seem to be right, the value is can't be right. When I checked the answer in the textbook it says it should be 14 /min.
Anyone have any input? 374