# 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 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 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 $$n = 1.48 , T = 20^{\circ} C , L = 0.03m ,\lambda_{0} = 589nm ,$$

$$\beta = 2.5*10^{-5} / ^{\circ} C , \alpha = 5.22*10^{-6} \frac{m}{ ^{\circ} C} , \Delta T = 5^{\circ} \frac{^{\circ} C}{min}$$

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?

• Linear expansion is in meters per meter-Celsius (i.e. does NOT have length units; you have incorrectly given its units in the delta-L equation. – Whit3rd Dec 8 '16 at 1:15
• Both your Delta equations are incorrect: the coefficient must be multiplied by the delta-T and by the original value... – DJohnM Dec 8 '16 at 4:45

At any temperature, the number of wavelengths of the light in the rod is given by:$$N_{T}=\frac{L\times n}{\lambda}$$where $L$ is the physical length, $n$ is the index of refraction, and $\lambda$ is the wavelength.
At the start, $T=20$, the number of wavelengths in the rod is given by:$$N_{20}=\frac{L_0\times n_o}{\lambda}$$After the temperature rises by 5 Celsius degrees, the number of waves increases; the rod has gotten slightly longer, and the waves of light have gotten shorter (or the optical length of the rod has increased; the math is the same)$$N_{25}=\frac{L_0\times (1+\alpha \Delta T)\times n_o \times(1+\beta \Delta T)}{\lambda}=\frac{L_0\times n_o}{\lambda}\times (1+(\alpha + \beta) \times\Delta T+\alpha \beta (\Delta T)^2)$$We can find the change in the number of waves by subtracting to get $$\Delta N=\frac{L_0\times n_o}{\lambda} \times (\alpha+ \beta )\Delta T$$The term in $(\Delta T)^2$ can be ignored; $\alpha +\beta$ is much bigger than $\alpha \times \beta$
One slight correction: since the rod has expanded, the free-air length of the arm has decreased; the rod has expanded into the space. The number of free-air waves removed is :$$N_{Rmvd}=\frac{L_O \times \alpha \times \Delta T}{\lambda}$$