Question

A disabled tanker leaks kerosene $\mathit{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{20}$into the Persian Gulf, creating a large slick on top of the water$\mathit{n}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{30}$). (a) If you are looking straight down from an airplane, while the Sun is overhead, at a region of the slick where its thickness is$\mathbf{460}\mathbf{}\mathbf{\text{nm}}$, for which wavelength(s) of visible light is the reflection brightest because of constructive interference? (b) If you are scuba diving directly under this same region of the slick, for which wavelength(s) of visible light is the transmitted intensity strongest?

Short answer

• The wavelength of the brightest reflected light is$552\text{nm}$
• The wavelength of the brightest transmitted light is $442\text{nm}$.

Step-by-step solution

Given data

• The refractive index of the kerosene is $n_2=1.20$.
• The refractive index of the water is$n_2=1.30$
• The thickness of the kerosene layer is $L=460\text{nm}$.

(a) Reflections from thin films.

Light incident normal on thin films reflects the light from its front and back surface resulting in interference of reflected lights. This interference gives bright reflected light when constructive interference occurs and dark spot when fully destructive interference occurs.

Here in this case light from the sun incident normal on the kerosene film. The refractive index of the kerosene film is higher than air so the reflected light from the front surface of the film will result in phase change. The second reflected light comes from the back surface of the layer, which goes through $180°$phase change. As a result, the condition for constructive interference is

role="math" localid="1663027768117" $2L=m\frac{\lambda }{n_2}$

Where $\lambda$is the wavelength of the light in air, $L$is its thickness, and $n_2$is the film’s refractive index.

Inserting the values from given data into the above equation to determine the wavelength of the brightest reflected light.

${\lambda }_{\max }=\frac{2L{n}_2}{m}$

role="math" localid="1663027910885" $\begin{array}{l}m=1;{\lambda }_1=\frac{2\left(460nm\right)\left(1.20\right)}{1}=1104\text{nm}\\ m=2;{\lambda }_2=\frac{2\left(460nm\right)\left(1.20\right)}{2}=552\text{nm}\\ m=3;{\lambda }_3=\frac{2\left(460nm\right)\left(1.20\right)}{3}=368\text{nm}\end{array}$

As $552\text{nm}$lies in visible range, hence the wavelength of the brightest reflected light is $552\text{nm}$.

(b) Transmission in thin films.

Interference of transmission line is similar to the interference of reflection of light. Here the phase difference between the transmitted rays $180°$is out of phase. This is because of the reflection off the back surface of the layer.

The condition for constructive interference is

$\begin{array}{c}2L=\left(m+\frac{1}{2}\right)\frac{\lambda }{n_2}\\ \lambda =\frac{4L{n}_2}{2m+1}\end{array}$

Calculating the wavelength for first few orders number,

$\begin{array}{l}m=0;{\lambda }_1=\frac{4\left(460nm\right)\left(1.20\right)}{2\left(0\right)+1}=2208\text{nm}\\ m=1;{\lambda }_2=\frac{4\left(460nm\right)\left(1.20\right)}{2\left(1\right)+1}=736\text{nm}\\ m=2;{\lambda }_3=\frac{4\left(460nm\right)\left(1.20\right)}{2\left(2\right)+1}=442\text{nm}\end{array}$

As $442\text{nm}$lies in visible range, hence the wavelength of the brightest transmitted light is $442\text{nm}$.