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?

  • $\begingroup$ 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. $\endgroup$
    – Whit3rd
    Commented Dec 8, 2016 at 1:15
  • $\begingroup$ Both your Delta equations are incorrect: the coefficient must be multiplied by the delta-T and by the original value... $\endgroup$
    – DJohnM
    Commented Dec 8, 2016 at 4:45

1 Answer 1


The glass rod makes up part of one arm of the Michelson Interferometer. So let's see how the number of wavelengths in the rod changes, and then correct for the reduced free-air length in the arm.

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 :$$N_{Rmvd}=\frac{L_O \times \alpha \times \Delta T}{\lambda}$$

Finally, since the light passes through the rod twice, the total wavelength shift is just double the net change above...

  • $\begingroup$ Thanks for the solution it helped a lot, not mention pointing out some of my silly mistakes. $\endgroup$
    – PatMac
    Commented Dec 8, 2016 at 11:14
  • $\begingroup$ I think we are not meant to present complete solutions to homework problems on this site. In the case of homework problems I think we are meant to offer just pointers. $\endgroup$ Commented Sep 30, 2020 at 9:07

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