Chapter 25, Problem 19P

(a)

To determine

The weakest wavelengths which are in the transmitted light.

Answer to Problem 19P

The weakest wavelengths in the transmitted light are $\underset{\_}{607\text{nm, 496}\text{nm, and 420}\text{nm}}$.

Explanation of Solution

Given that the index of refraction of the soap film is $1.50$, the thickness of the film is $910.0\text{nm}$.

The weakest wavelengths in transmitted light correspond to the strongest wavelengths in the reflected light. Rays which are found as reflected off the front of the film will be reversed in phase, since the refractive index of the air is found to be less than the refractive index of the film.

However, the rays reflected off the back of the film are not inverted and thus the reflected rays will be out of phase and the $180°$ plus the phase shift caused by the path length difference.

The condition for constructive interference is used to find the strongest wavelengths in reflected light. This occurs when the path length difference, $2t$ is an odd multiple of half the wavelength in the film.

Write the condition for the constructive interference.

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

Here, $t$ is the thickness of the film, $m$ is the order of interference, $\lambda$ is the wavelength of the transmitted light.

Write the expression for the wavelength of the transmitted light.

    $\lambda =\frac{{\lambda }_0}{n_{\text{film}}}$                                                                                                                   (II)

Here, ${\lambda }_0$ is the weakest wavelengths in transmitted light, $n_{\text{film}}$ is the refractive index of the film.

Use equation (II) in equation (I),

    $2t=\left(m+\frac{1}{2}\right)\frac{{\lambda }_0}{n_{\text{film}}}$                                                                                                   (III)

Solve equation (III) for ${\lambda }_0$,

    ${\lambda }_0=\frac{2t{n}_{\text{film}}}{\left(m+\frac{1}{2}\right)}$                                                                                                         (IV)

Conclusion:

Substitute $910.0\text{nm}$ for $t$, $1.50$ for $n_{\text{film}}$ and 4 for $m$ in equation (IV) to find ${\lambda }_0$ corresponding to 4^th order.

    $\begin{array}{c}{\lambda }_0=\frac{2\left(910.0\text{nm}\right)\left(\text{1}\text{.50}\right)}{4+\frac{1}{2}}\\ =607\text{nm}\end{array}$

Substitute $910.0\text{nm}$ for $t$, $1.50$ for $n_{\text{film}}$ and 5 for $m$ in equation (IV) to find ${\lambda }_0$ corresponding to 5^th order.

    $\begin{array}{c}{\lambda }_0=\frac{2\left(910.0\text{nm}\right)\left(\text{1}\text{.50}\right)}{5+\frac{1}{2}}\\ =496\text{nm}\end{array}$

Substitute $910.0\text{nm}$ for $t$, $1.50$ for $n_{\text{film}}$ and 6 for $m$ in equation (IV) to find ${\lambda }_0$ corresponding to 6^th order.

    $\begin{array}{c}{\lambda }_0=\frac{2\left(910.0\text{nm}\right)\left(\text{1}\text{.50}\right)}{6+\frac{1}{2}}\\ =420\text{nm}\end{array}$

Therefore, the weakest wavelengths in the transmitted light are $\underset{\_}{607\text{nm, 496}\text{nm, and 420}\text{nm}}$.

(b)

To determine

Wavelengths which are strongest in the transmitted light.

Answer to Problem 19P

The wavelengths which are strongest in the transmitted light are $\underset{\_}{683\text{nm, 546}\text{nm, 455}\text{nm}}$.

Explanation of Solution

The strongest wavelengths in transmitted light correspond to the weakest wavelengths in reflected light. Destructive interference in the reflected light occurs when the path difference, $2t$ is equal to an integral number of wavelengths in the film.

Write the condition for the destructive interference.

    $2t=m\lambda$                                                                                                                   (V)

Use equation (II) in equation (I).

    $2t=m\frac{{\lambda }_0}{n_{\text{film}}}$                                                                                                             (VI)

Solve equation (VI) for ${\lambda }_0$.

    ${\lambda }_0=\frac{2t{n}_{\text{film}}}{m}$                                                                                                            (VII)

Conclusion:

Substitute $910.0\text{nm}$ for $t$, $1.50$ for $n_{\text{film}}$ and 4 for $m$ in equation (VII) to find ${\lambda }_0$ corresponding to 4^th order.

    $\begin{array}{c}{\lambda }_0=\frac{2\left(910.0\text{nm}\right)\left(\text{1}\text{.50}\right)}{4}\\ =683\text{nm}\end{array}$

Substitute $910.0\text{nm}$ for $t$, $1.50$ for $n_{\text{film}}$ and 5 for $m$ in equation (VII) to find ${\lambda }_0$ corresponding to 5^th order.

    $\begin{array}{c}{\lambda }_0=\frac{2\left(910.0\text{nm}\right)\left(\text{1}\text{.50}\right)}{5}\\ =546\text{nm}\end{array}$

Substitute $910.0\text{nm}$ for $t$, $1.50$ for $n_{\text{film}}$ and 6 for $m$ in equation (VII)  to find ${\lambda }_0$ corresponding to 6^th order.

    $\begin{array}{c}{\lambda }_0=\frac{2\left(910.0\text{nm}\right)\left(\text{1}\text{.50}\right)}{6}\\ =455\text{nm}\end{array}$

Therefore, the wavelengths which are strongest in the transmitted light are $\underset{\_}{683\text{nm, 546}\text{nm, 455}\text{nm}}$.