91影视

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

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

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,

$\mathit{螖}{\mathit{n}}_{\mathbf{r}\mathbf{o}\mathbf{d}}\mathbf{=}{\mathit{n}}_{\mathbf{i}\mathbf{,}\mathbf{r}\mathbf{o}\mathbf{d}}{\mathit{伪}}_{\mathbf{n}}\mathit{螖}\mathit{T}$

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

So,

$\mathit{螖}{\mathit{L}}_{\mathbf{r}\mathbf{o}\mathbf{d}}\mathbf{=}{\mathit{L}}_{\mathbf{i}}{\mathit{伪}}_{\mathbf{r}\mathbf{o}\mathbf{d}}\mathit{螖}\mathit{T}$

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

$螖N=\left(2螖L_{rod}\right)/{位}_{rod}$

From snell鈥檚 law we have

$n_1{位}_1=n_2{位}_2$

And here we get

$n_{air}{位}_{air}=n_{rod}{位}_{rod}$

Solving forand=1

${位}_{rod}=\frac{{位}_{air}}{n_{rod}}$鈥�

Input this in the formula

$螖N=\frac{2n_{rod}螖L_{rod}}{{位}_{air}}$

We need the number of fringes each minute so

$\frac{螖N}{螖t}=\frac{2n_{rod}螖L_{rod}}{{位}_{air}螖t}$

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

$\frac{螖N_{air}}{螖t}=\frac{2n_{air}螖L_{rod}}{{位}_{air}螖t}$

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

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

Plugging in the values

$\frac{螖N_{tot}}{螖t}=\frac{2n_{rod}鈭�L_{rod}}{{位}_{air}螖t}+\frac{{2_n}_{air}鈭�L_{rod}}{{位}_{air}螖t}+\frac{2螖n_{rod}{L}_i}{{位}_{air}螖t}$

Substitute $鈭�L_{rod}$

$\frac{螖N_{tot}}{螖t}=(n_{rod}-n_{air})\frac{2L_i{伪}_{rod}螖T}{{位}_{air}螖t}+\frac{2螖n_{rod}{L}_i}{{位}_{air}螖t}$

Substitute $鈭�n_{rod}$

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

Rearrange to get

$\frac{螖N_{tot}}{螖t}=\left(\right(n_{rod}-n_{air})\frac{2L_i{伪}_{rod}}{{位}_{air}}+\frac{2_{i,rod}{伪}_n{L}_i}{{位}_{air}})\frac{鈭�T}{鈭�t}$

Input the values

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

Solving we get

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

Hence, we see 14fringes per minute.