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 theglass is 5.00* 10-6/C°. 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 l=589 nm, and the rod initially is at T=20.0°C. How many fringes cross the field of view each minute?

Constructive interference is at $\mathit{n}\mathit{\lambda }$

and destructive interference is at$\frac{\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{\lambda }}{\mathbf{2}}$.

Where n is an integer.

The index of refraction of the rod varies linearly with temperature,and we are given the linear coefficient of this change

So,

$\mathit{\Delta }{\mathit{n}}_{\mathbf{r}\mathbf{o}\mathbf{d}}\mathbf{=}{\mathit{n}}_{\mathbf{i}\mathbf{,}\mathbf{r}\mathbf{o}\mathbf{d}}{\mathit{\alpha }}_{\mathbf{n}}\mathit{\Delta }\mathit{T}$

We also know that the length of the rod linearly varies with temperature

So,

$\mathit{\Delta }{\mathit{L}}_{\mathbf{r}\mathbf{o}\mathbf{d}}\mathbf{=}{\mathit{L}}_{\mathbf{i}}{\mathit{\alpha }}_{\mathbf{r}\mathbf{o}\mathbf{d}}\mathit{\Delta }\mathit{T}$

Noting that the change in the number of the fringes is caused by the change of the rod length due to heating expansion

$\Delta N=\left(2\Delta L_{rod}\right)/{\lambda }_{rod}$

From snell’s law we have

$n_1{\lambda }_1=n_2{\lambda }_2$

And here we get

$n_{air}{\lambda }_{air}=n_{rod}{\lambda }_{rod}$

Solving forand=1

${\lambda }_{rod}=\frac{{\lambda }_{air}}{n_{rod}}$’

Input this in the formula

$\Delta N=\frac{2n_{rod}\Delta L_{rod}}{{\lambda }_{air}}$

We need the number of fringes each minute so

$\frac{\Delta N}{\Delta t}=\frac{2n_{rod}\Delta L_{rod}}{{\lambda }_{air}\Delta t}$

The change in the number of fringes is due to air, is given by, using the same approach

$\frac{\Delta N_{air}}{\Delta t}=\frac{2n_{air}\Delta L_{rod}}{{\lambda }_{air}\Delta t}$

The minus sign is due to the decrease of the air due to increase of the length of the rod

Hence the total change is given by

$\frac{\Delta N_{tot}}{\Delta t}=\frac{\Delta N_{rod}}{\Delta t}+\frac{\Delta N_{air}ir}{\Delta t}+\frac{\Delta {N_n}_{rod}}{\Delta t}$

Plugging in the values

$\frac{\Delta N_{tot}}{\Delta t}=\frac{2n_{rod}∆L_{rod}}{{\lambda }_{air}\Delta t}+\frac{{2_n}_{air}∆L_{rod}}{{\lambda }_{air}\Delta t}+\frac{2\Delta n_{rod}{L}_i}{{\lambda }_{air}\Delta t}$

Substitute $∆L_{rod}$

$\frac{\Delta N_{tot}}{\Delta t}=(n_{rod}-n_{air})\frac{2L_i{\alpha }_{rod}\Delta T}{{\lambda }_{air}\Delta t}+\frac{2\Delta n_{rod}{L}_i}{{\lambda }_{air}\Delta t}$

Substitute $∆n_{rod}$

$\frac{\Delta N_{tot}}{\Delta t}=(n_{rod}-n_{air})\frac{2L_i{\alpha }_{rod}\Delta T}{{\lambda }_{air}\Delta t}+\frac{2{n_{i,}}_{rod}{\alpha }_n{L}_i\Delta T}{{\lambda }_{air}\Delta t}$

Rearrange to get

$\frac{\Delta N_{tot}}{\Delta t}=\left(\right(n_{rod}-n_{air})\frac{2L_i{\alpha }_{rod}}{{\lambda }_{air}}+\frac{2_{i,rod}{\alpha }_n{L}_i}{{\lambda }_{air}})\frac{∆T}{∆t}$

Input the values

$\frac{\Delta N_{tot}}{\Delta t}=\left(\right(1.48-1.0)\frac{2\times 3.0\times {10}^{-6}\times 5.0\times {10}^{-6}}{583\times {10}^{-9}}+\frac{2\times 2.5\times {10}^{-6}\times 3.0\times {10}^{-6}}{583\times {10}^{-9}})\frac{5.0}{6.0}$

Solving we get

$$\begin{gathered} \frac{\Delta N_{tot}}{\Delta t}=0.2326fringe/s \\ \frac{\Delta N_{tot}}{\Delta t}=14.0fringe/min \end{gathered}$$

Hence, we see 14fringes per minute.