Chapter 37, Problem 54AP

To determine

The coefficient of the linear expansion of metal.

Answer to Problem 54AP

The coefficient of the linear expansion of metal is $\text{2}\text{.5}\times {\text{10}}^{-6}{\text{C}}^{-1}$.

Explanation of Solution

Given info: The initial length of the metal is $10.0\text{cm}$, the wavelength of the light is $500\text{nm}$, the change of the order of fringes is $200$ and the refractive index for air and the increased temperature is $20.0°$

The condition for the constructive interference is,

$2nt=\left(m+\frac{1}{2}\right)\lambda$ (1)

Here,

$n$ is the refractive index for air.

$m$ is the order of the fringes.

$\lambda$ is the wavelength of the light.

$t$ is the thickness of the film.

Substitute $1$ for $n$ in equation (1).

$\begin{array}{c}2\times 1\times t=\left(m+\frac{1}{2}\right)\lambda \\ 2t=\left(m+\frac{1}{2}\right)\lambda \end{array}$ (2)

Rearrange the equation (2) to find the $t$.

$t=\left(m+\frac{1}{2}\right)\frac{\lambda }{2}$

The initial thickness of the film is,

$t_{\text{i}}=\left(m_{\text{i}}+\frac{1}{2}\right)\frac{\lambda }{2}$

Here,

$t_{\text{i}}$ is the initial thickness of metal.

$m_{\text{i}}$ is the initial order of fringe.

The final thickness of the film is,

$t_{\text{f}}=\left(m_{\text{f}}+\frac{1}{2}\right)\frac{\lambda }{2}$

Here,

$t_{\text{f}}$ is the final thickness of metal

$m_{\text{f}}$ is the final order of fringe.

Since, the temperature slowly increases, the rod undergoes an expansion in the length and the thickness of the air decreases.

$\Delta l=t_{\text{i}}-t_{\text{f}}$ (3)

Here,

$\Delta l$ is the change of the length of the metal.

Substitute $\left(m_{\text{f}}+\frac{1}{2}\right)\frac{\lambda }{2}$ for $t_{\text{f}}$ and $\left(m_{\text{i}}+\frac{1}{2}\right)\frac{\lambda }{2}$ for $t_{\text{i}}$ in equation (3).

$\begin{array}{c}\Delta l=\left(m_{\text{f}}+\frac{1}{2}\right)\frac{\lambda }{2}-\left(m_{\text{i}}+\frac{1}{2}\right)\frac{\lambda }{2}\\ =\left(m_{\text{f}}-m_{\text{i}}\right)\frac{\lambda }{2}\end{array}$ (4)

The change of the order of fringes is,

$\Delta m=m_{\text{i}}-m_{\text{f}}$

Here,

$\Delta m$ is the change of the order of fringes.

Substitute $m_{\text{i}}-m_{\text{f}}$ for $\Delta m$ in equation (4).

$\begin{array}{c}\Delta L=\left(m_{\text{f}}+\frac{1}{2}\right)\frac{\lambda }{2}-\left(m_{\text{i}}+\frac{1}{2}\right)\frac{\lambda }{2}\\ =\left(\Delta m\right)\frac{\lambda }{2}\end{array}$

Formula to calculate the coefficient of linear expansion of the material is,

$\alpha =\frac{\Delta L}{L_{\text{i}}\Delta T}$ (5)

Here,

$\alpha$ is the coefficient of linear expansion of the material.

$L_{\text{i}}$ is the initial length of the metal.

$\Delta T$ is the changed temperature.

Substitute $\left(\Delta m\right)\frac{\lambda }{2}$ for $\Delta L$ in equation (5).

$\alpha =\frac{\left(\Delta m\right)\frac{\lambda }{2}}{L_{\text{i}}\Delta T}$ (6)

The change of the temperature is,

$\Delta T=T_{\text{i}}-T_{\text{f}}$ (7)

Substitute $25.0°\text{C}$ for $T_{\text{f}}$ and $0.0°\text{C}$ for $T_{\text{i}}$ in equation (7) to find the $\Delta T$.

$\begin{array}{c}\Delta T=25.0°\text{C}-0.0°\text{C}\\ \text{=}25.0°\text{C}\end{array}$

Substitute $200$ for $\Delta m$, $500\text{nm}$ for $\lambda$, $10.0\text{cm}$ for $L_{\text{i}}$ and $25.0°\text{C}$ for $\Delta T$ in equation (6).

$\begin{array}{l}\alpha =\frac{200\frac{\left(500\text{nm}\left(\frac{{10}^{-9}\text{m}}{\text{1}\text{nm}}\right)\right)}{2}}{10.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)25.0°\text{C}}\\ \text{=2}\times {\text{10}}^{-6}{\text{C}}^{-1}\end{array}$

Conclusion:

Therefore, the coefficient of the linear expansion of metal $\text{2}\text{.5}\times {\text{10}}^{-6}{\text{C}}^{-1}$.