Chapter 25, Problem 81P

(a)

To determine

The thickness of the soap film.

Answer to Problem 81P

The thickness of the soap film is $\underset{\_}{1.11\text{μm}}$.

Explanation of Solution

The ray reflected from the front of the film will be reversed in phase because the refractive index of air is less than that of the film.

 The ray reflected from the back of the film will not be inverted since the film has higher refractive index. The two reflected rays are $180°$ out of phase, plus the phase shift caused by the path length difference $2t$ in the film.

Missing wavelengths in the reflected light are caused by destructive interference, which occurs when the path length difference is equal to an integral multiple of the wavelength in the film. The reflections in the film are shown in Figure 1.

Write the condition of path length difference in the system for normal incidence.

  $2t=m\lambda$                                                                                                                   (I)

Here, $t$ is the thickness of the film, $m$ is the order of diffraction, and $\lambda$ is the wavelength of light in the film.

Write the expression for the wavelength of light in the film.

  $\lambda =\frac{{\lambda }_0}{n}$                                                                                                                    (II)

Here, ${\lambda }_0$ is the wavelength of length of light in vacuum, and $n$ is the refractive index of the film.

Use equation (II) in (I).

  $\begin{array}{c}2t=m\frac{{\lambda }_0}{n}\\ 2tn=m{\lambda }_0\end{array}$                                                                                                             (III)

Since $2tn$ is a constant, for any two missing wavelengths ${\lambda }_1$ and ${\lambda }_2$ and their respective order of interference, it can be written that,

  $m_1{\lambda }_1=m_2{\lambda }_2$                                                                                                            (IV)

Use $m_2=m_1+1$ in equation (IV) and solve for $m_1$.

  $\begin{array}{c}{m}_1{\lambda }_1=\left(m_1+1\right){\lambda }_2\\ m_1=\frac{{\lambda }_2}{{\lambda }_1-{\lambda }_2}\end{array}$                                                                                                    (V)

Replace $m$ and ${\lambda }_0$ in equation (III) by $m_1$ and ${\lambda }_1$ respectively, and solve for $t$.

  $t=\frac{m_1{\lambda }_1}{2n}$                                                                                                                 (VI)

Conclusion:

Substitute $600.0\text{nm}$ for ${\lambda }_1$, and $500.0\text{nm}$ for ${\lambda }_2$ in equation (V) to find $m_1$.

  $\begin{array}{c}{m}_1=\frac{500.0\text{nm}}{600.0\text{nm}-500.0\text{nm}}\\ =5.000\end{array}$

Substitute $600.0\text{nm}$ for ${\lambda }_1$, $5.000$ for $m_1$, and $1.35$ for $n$ in equation (VI) to find $t$.

  $\begin{array}{c}t=\frac{\left(5.000\right)\left(600.0\text{nm}\right)}{2\left(1.35\right)}\\ =\frac{\left(5.000\right)\left(600.0\text{nm}\times \frac{1\text{μm}}{1\times {10}^3\text{nm}}\right)}{2\left(1.35\right)}\\ =1.11\text{μm}\end{array}$

Therefore, the thickness of the soap film is $\underset{\_}{1.11\text{μm}}$.

(b)

To determine

Whether there are any other missing visible wavelengths from the reflected light, and their wavelengths.

Answer to Problem 81P

There is missing visible wavelength in the reflected light and it is $\underset{\_}{429\text{nm}}$.

Explanation of Solution

The missing wavelengths of $600.0\text{nm}$ and $500.0\text{nm}$ correspond to $m=5$ and $m=6$ respectively. In order to see if other wavelengths are missing, the destructive interference equation (III) has to be solved for ${\lambda }_0$.

  ${\lambda }_m=\frac{2tn}{m}$                                                                                                              (VII)

Conclusion:

Substitute $1.11\text{μm}$ for $t$, $1.35$ for $n$, and $4$ for $m$ in equation (VII) to find ${\lambda }_m$ corresponding to the $4^{\text{th}}$ order spectrum.

  $\begin{array}{c}{\lambda }_4=\frac{2\left(1.11\text{μm}\right)\left(1.35\right)}{4}\\ =\frac{2\left(1.11\text{μm}\times \frac{1000\text{nm}}{1\text{μm}}\right)\left(1.35\right)}{4}\\ =750\text{nm}\end{array}$

Substitute $1.11\text{μm}$ for $t$, $1.35$ for $n$, and $7$ for $m$ in equation (VII) to find ${\lambda }_m$ corresponding to the $7^{\text{th}}$ order spectrum.

  $\begin{array}{c}{\lambda }_7=\frac{2\left(1.11\text{μm}\right)\left(1.35\right)}{7}\\ =\frac{2\left(1.11\text{μm}\times \frac{1000\text{nm}}{1\text{μm}}\right)\left(1.35\right)}{7}\\ =429\text{nm}\end{array}$

Substitute $1.11\text{μm}$ for $t$, $1.35$ for $n$, and $8$ for $m$ in equation (VII) to find ${\lambda }_m$ corresponding to the $8^{\text{th}}$ order spectrum.

  $\begin{array}{c}{\lambda }_7=\frac{2\left(1.11\text{μm}\right)\left(1.35\right)}{8}\\ =\frac{2\left(1.11\text{μm}\times \frac{1000\text{nm}}{1\text{μm}}\right)\left(1.35\right)}{8}\\ =375\text{nm}\end{array}$

Among the obtained values of the missing wavelengths, the wavelength of $429\text{nm}$ only lies in the visible region.

Therefore, there is missing visible wavelength in the reflected light and it is $\underset{\_}{429\text{nm}}$.

(c)

To determine

The wavelengths of light that are strongest in the transmitted light.

Answer to Problem 81P

The wavelengths of light that are strongest in the transmitted light are $\underset{\_}{429\text{nm, }500.0\text{nm}}$, and $\underset{\_}{600.0\text{nm}}$.

Explanation of Solution

The wavelengths that are strongest in transmitted light are the wavelengths that are missing in the reflected light. For the visible region of the electromagnetic spectrum, these are $429\text{nm}$, $500.0\text{nm}$, and $600.0\text{nm}$.

Conclusion:

Therefore, the wavelengths of light that are strongest in the transmitted light are $\underset{\_}{429\text{nm, }500.0\text{nm}}$, and $\underset{\_}{600.0\text{nm}}$.