91Ó°ÊÓ

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

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

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,

$\mathit{Δ}{\mathit{n}}_{\mathbf{r}\mathbf{o}\mathbf{d}}\mathbf{=}{\mathit{n}}_{\mathbf{i}\mathbf{,}\mathbf{r}\mathbf{o}\mathbf{d}}{\mathit{Î\pm }}_{\mathbf{n}}\mathit{Δ}\mathit{T}$

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

So,

$\mathit{Δ}{\mathit{L}}_{\mathbf{r}\mathbf{o}\mathbf{d}}\mathbf{=}{\mathit{L}}_{\mathbf{i}}{\mathit{Î\pm }}_{\mathbf{r}\mathbf{o}\mathbf{d}}\mathit{Δ}\mathit{T}$

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

$\mathrm{Δ}N=\left(2\mathrm{Δ}{L}_{rod}\right)/{\mathrm{λ}}_{rod}$

From snell’s law we have

$n_1{\mathrm{λ}}_1=n_2{\mathrm{λ}}_2$

And here we get

$n_{air}{\mathrm{λ}}_{air}=n_{rod}{\mathrm{λ}}_{rod}$

Solving forand=1

${\mathrm{λ}}_{rod}=\frac{{\mathrm{λ}}_{air}}{n_{rod}}$’

Input this in the formula

$\mathrm{Δ}N=\frac{2n_{rod}\mathrm{Δ}{L}_{rod}}{{\mathrm{λ}}_{air}}$

We need the number of fringes each minute so

$\frac{\mathrm{Δ}N}{\mathrm{Δ}t}=\frac{2n_{rod}\mathrm{Δ}{L}_{rod}}{{\mathrm{λ}}_{air}\mathrm{Δ}t}$

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

$\frac{\mathrm{Δ}{N}_{air}}{\mathrm{Δ}t}=\frac{2n_{air}\mathrm{Δ}{L}_{rod}}{{\mathrm{λ}}_{air}\mathrm{Δ}t}$

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

$\frac{\mathrm{Δ}{N}_{tot}}{\mathrm{Δ}t}=\frac{\mathrm{Δ}{N}_{rod}}{\mathrm{Δ}t}+\frac{\mathrm{Δ}{N}_{air}ir}{\mathrm{Δ}t}+\frac{\mathrm{Δ}{N_n}_{rod}}{\mathrm{Δ}t}$

Plugging in the values

$\frac{\mathrm{Δ}{N}_{tot}}{\mathrm{Δ}t}=\frac{2n_{rod}∆L_{rod}}{{\mathrm{λ}}_{air}\mathrm{Δ}t}+\frac{{2_n}_{air}∆L_{rod}}{{\mathrm{λ}}_{air}\mathrm{Δ}t}+\frac{2\mathrm{Δ}{n}_{rod}{L}_i}{{\mathrm{λ}}_{air}\mathrm{Δ}t}$

Substitute $∆L_{rod}$

$\frac{\mathrm{Δ}{N}_{tot}}{\mathrm{Δ}t}=(n_{rod}-n_{air})\frac{2L_i{\mathrm{Î\pm }}_{rod}\mathrm{Δ}T}{{\mathrm{λ}}_{air}\mathrm{Δ}t}+\frac{2\mathrm{Δ}{n}_{rod}{L}_i}{{\mathrm{λ}}_{air}\mathrm{Δ}t}$

Substitute $∆n_{rod}$

$\frac{\mathrm{Δ}{N}_{tot}}{\mathrm{Δ}t}=(n_{rod}-n_{air})\frac{2L_i{\mathrm{Î\pm }}_{rod}\mathrm{Δ}T}{{\mathrm{λ}}_{air}\mathrm{Δ}t}+\frac{2{n_{i,}}_{rod}{\mathrm{Î\pm }}_n{L}_i\mathrm{Δ}T}{{\mathrm{λ}}_{air}\mathrm{Δ}t}$

Rearrange to get

$\frac{\mathrm{Δ}{N}_{tot}}{\mathrm{Δ}t}=\left(\right(n_{rod}-n_{air})\frac{2L_i{\mathrm{Î\pm }}_{rod}}{{\mathrm{λ}}_{air}}+\frac{2_{i,rod}{\mathrm{Î\pm }}_n{L}_i}{{\mathrm{λ}}_{air}})\frac{∆T}{∆t}$

Input the values

$\frac{\mathrm{Δ}{N}_{tot}}{\mathrm{Δ}t}=\left(\right(1.48-1.0)\frac{2×3.0×{10}^{-6}×5.0×{10}^{-6}}{583×{10}^{-9}}+\frac{2×2.5×{10}^{-6}×3.0×{10}^{-6}}{583×{10}^{-9}})\frac{5.0}{6.0}$

Solving we get

$$\begin{gathered} \frac{\mathrm{Δ}{N}_{tot}}{\mathrm{Δ}t}=0.2326fringe/s \\ \frac{\mathrm{Δ}{N}_{tot}}{\mathrm{Δ}t}=14.0fringe/min \end{gathered}$$

Hence, we see 14fringes per minute.