Chapter 40, Problem 32SP

(a)

To determine

The first three smallest thicknesses of the film for the constructive interference when wavelength of the wave in vacuum is $600\text{ nm}$ and the refractive index of the film is $1.40$.

Answer to Problem 32SP

Solution:

$0\text{ nm}$, $214.3\text{ nm}$, and $429\text{ nm}$

Explanation of Solution

Given data:

The wavelength of the wave in vacuum is $600\text{ nm}$.

The refractive index of the film is $1.40$.

Formula used:

The expression for the path length difference for the constructive interference of the waves is

$\Delta r=m{\lambda }_0$

Here, the values of $m$ varies as $\left(0,1,2,3,4,\mathrm{......}\right)$, ${\lambda }_0$ is the wavelength of the monochromatic light, and $\Delta r$ is the path difference of the wave from both the sources.

Explanation:

Refer to the Fig. 40-2 provided in the textbook:

Write the expression for the path length difference of the reflected waves, from upper and the lower surface of the film, for the constructive interference:

$\Delta r=m{\lambda }_0$

The wave reflected by the lower and the upper surface of the film has the path difference, which is equals to $2nd$.

Substitute $2nd$ for $\Delta r$

$2nd=m{\lambda }_0$

Here, $n$ is the refractive index of the film and $d$ is the thickness of the film.

Rearrange for $d$

$d=\frac{m\lambda }{2n}$

Substitute $600\text{ nm }$ for ${\lambda }_0$ and $1.40$ for $n$

$d=\frac{m\left(600\text{ nm }\right)}{2\left(1.40\right)}$

For $m=0$

$\begin{array}{c}d=\frac{\left(0\right)\left(600\text{ nm }\right)}{2\left(1.40\right)}\\ =0\text{ nm}\end{array}$

For $m=1$

$\begin{array}{c}d=\frac{\left(1\right)\left(600\text{ nm }\right)}{2\left(1.40\right)}\\ =214.3\text{ nm}\\ \approx \text{214 nm}\end{array}$

For $m=2$.

$\begin{array}{c}d=\frac{\left(2\right)\left(600\text{ nm }\right)}{2\left(1.40\right)}\\ =429\text{ nm}\end{array}$

Conclusion:

The thicknesses of the film for the constructive interference are $0\text{ nm}$, $214\text{ nm}$, and $429\text{ nm}$.

(b)

To determine

The first three smallest thicknesses of the film for the destructive interference when the refractive index of the film is given.

Answer to Problem 32SP

Solution:

$107\text{ nm}$, $321\text{ nm}$, and $536\text{ nm}$

Explanation of Solution

Formula used:

The expression for the path length difference for the destructive interference of the waves is

$\Delta r=\left(m+\frac{1}{2}\right){\lambda }_0$

Here, the values of $m$ varies as $\left(0,1,2,3,4,\mathrm{......}\right)$, ${\lambda }_0$ is the wavelength of the monochromatic light, and $\Delta r$ is the path difference between the waves.

Explanation:

Write the expression for the path length difference of the reflected waves, from upper and the lower surface of the film, for the destructive interferences:

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

The wave reflected by the lower and upper surface of the film has the optical path length difference of $2nd$.

Substitute $2nd$ for $\Delta r$ and $600\text{ nm}$ for $\lambda$

$2nd=\left(m+\frac{1}{2}\right)600\text{ nm }$

Rearrange for $d$

$d=\frac{\left(m+\frac{1}{2}\right)600\text{ nm }}{2n}$

Substitute $1.40$ for $n$

$\begin{array}{c}d=\frac{\left(m+\frac{1}{2}\right)600\text{ nm }}{2\left(1.40\right)}\\ =\left(2m+1\right)\text{107}\text{.14 nm}\end{array}$

For $m=0$

$\begin{array}{c}d=\left(2\left(0\right)+1\right)\text{107}\text{.14 nm}\\ =\text{107}\text{.14 nm}\\ \simeq \text{107 nm}\end{array}$

For $m=1$

$\begin{array}{c}d=\left(2\left(1\right)+1\right)\text{107}\text{.14 nm}\\ =\text{321}\text{.43 nm}\\ \approx \text{ 321 nm}\end{array}$

For $m=2$

$\begin{array}{c}d=\left(2\left(2\right)+1\right)\text{107}\text{.14 nm}\\ =\text{536 nm}\end{array}$

Conclusion:

The thicknesses of the film for the constructive interference are $107\text{ nm}$, $321\text{ nm}$, and $536\text{ nm}$.