Chapter 36, Problem 49CP

(a)

To determine

The expression of the integer $m$ in terms of the wavelength of ${\lambda }_1$ and ${\lambda }_2$.

Answer to Problem 49CP

The expression of the integer $m$ in terms of the wavelength of ${\lambda }_1$ and ${\lambda }_2$ is $\frac{{\lambda }_1}{2\left({\lambda }_1-{\lambda }_2\right)}$.

Explanation of Solution

Given info: The index reflection for film is $n$, the thickness of the slit is $d$, maximum intensity is observed at ${\lambda }_1$ and minimum intensity is observed at ${\lambda }_2$.

The condition of the destructive interference for minimum intensity is,

$2nt=m{\lambda }_2$ (1)

Here,

$m$ is the change of order.

${\lambda }_2$ is the wavelength for the minimum intensity.

${\lambda }_1$ is the wavelength for the maximum intensity.

$t$ is the thickness of the film.

Rearrange the equation (1) to find the ${\lambda }_2$.

${\lambda }_2=\frac{2nt}{m}$

The condition of constructive interference for the maximum intensity of fringes is,

$2nt=\left(m+\frac{1}{2}\right){\lambda }_1$ (2)

Rearrange the equation (2) to find the ${\lambda }_1$.

${\lambda }_1=\frac{2nt}{\left(m\text{'}+\frac{1}{2}\right)}$

Since, $m$ and $m\text{'}$ are distinct integer values and must be consecutive because no intensity minima are observed between ${\lambda }_1$ and ${\lambda }_2$.

The wavelength of the maximum intensity is greater than the wavelength of the minimum intensity.

${\lambda }_1>{\lambda }_2$

Substitute $\frac{2nt}{\left(m\text{'}+\frac{1}{2}\right)}$ for ${\lambda }_1$ and $\frac{2nt}{m}$ for ${\lambda }_2$.

$\begin{array}{c}\frac{2nt}{\left(m\text{'}+\frac{1}{2}\right)}>\frac{2nt}{m}\\ \left(m\text{'}+\frac{1}{2}\right)<m\end{array}$

Hence, the above equation exists for $m\text{'}=m-1$.

Substitute $\left(m\text{'}+\frac{1}{2}\right)$ for $m$ in equation (1)

$2nt=\left(m\text{'}+\frac{1}{2}\right){\lambda }_2$ (3)

Substitute $m-1$ for $m\text{'}$ in equation (3) to find the $m$.

$\begin{array}{c}2nt=\left(m-1+\frac{1}{2}\right){\lambda }_2\\ m=\frac{{\lambda }_1}{2\left({\lambda }_1-{\lambda }_2\right)}\end{array}$

Conclusion:

Therefore, the expression of the integer $m$ in terms of the wavelength of ${\lambda }_1$ and ${\lambda }_2$ is $\frac{{\lambda }_1}{2\left({\lambda }_1-{\lambda }_2\right)}$.

(b)

To determine

The best thickness of the film.

Answer to Problem 49CP

The best thickness of the film is $266\text{nm}$.

Explanation of Solution

Given info: The index reflection for film is $1.40$, maximum intensity is observed at $500\text{nm}$ and minimum intensity is observed at $\text{370}\text{nm}$.

Thus, the expression of the integer order of fringe $m$ is and is,

$m=\frac{{\lambda }_1}{2\left({\lambda }_1-{\lambda }_2\right)}$

Substitute $500\text{nm}$ for ${\lambda }_1$ and $\text{370}\text{nm}$ for ${\lambda }_2$ to find the $m$.

$\begin{array}{c}m=\frac{500\text{nm}}{2\left(500\text{nm}-370\text{nm}\right)}\\ =1.92\\ \approx 2\end{array}$

The condition of the destructive interference for minimum intensity is,

$2nt=m{\lambda }_2$ (4)

Substitute $1.40$ for $n$, $2$ for $m$ and $\text{370}\text{nm}$ for ${\lambda }_2$ in equation (4) to find the $t$.

$\begin{array}{c}2\left(1.40\right)t=2\left(\text{370}\text{nm}\right)\\ t=264.28\text{nm}\\ \approx 264.28\text{nm}\end{array}$

The condition of constructive interference for the maximum intensity of fringes is,

$2nt=\left(m\text{'}+\frac{1}{2}\right){\lambda }_1$ (5)

Substitute $m-1$ for $m\text{'}$ in equation (5).

$\begin{array}{c}2nt=\left(m-1+\frac{1}{2}\right){\lambda }_2\\ 2nt=\left(m-\frac{1}{2}\right){\lambda }_2\end{array}$ (6)

Substitute $1.40$ for $n$, $2$ for $m$ and $\text{370}\text{nm}$ for ${\lambda }_2$ in equation (4) to find the $t$

$\begin{array}{c}2\left(1.40\right)t=\left(2-\frac{1}{2}\right)\left(\text{500}\text{nm}\right)\\ t=267.85\text{nm}\\ \approx 368\text{nm}\end{array}$

The average of the thickness at minimum intensity and thickness of the maximum intensity is,

$\frac{268\text{nm}+264\text{nm}}{2}=266\text{nm}$

The average of the thickness at minimum intensity and thickness of the maximum intensity represents the best thickness of film.

Conclusion:

Therefore, the best thickness of the film is $266\text{nm}$.