Question

A scientist notices that an oil slick floating on water when viewed from above has many different rainbow colors reflecting off the surface. She aims a spectrometer at a particular spot and measures the wavelength to be 750 nm (in air). The index of refraction of water is 1.33.

Part A: The index of refraction of the oil is 1.20. What is the minimum thickness of the oil slick at that spot? t= 313nm

Part B: Suppose the oil had an index of refraction of 1.50. What would the minimum thickness be now? t=125nm

I only need help with part c. I put this in incase you need the information.

Part C: Now assume that the oil had a thickness of 200 nm and an index of refraction of 1.5. A diver swimming underneath the oil slick is looking at the same spot as the scientist with the spectromenter. What is the longest wavelength of the light in water that is transmitted most easily to the diver?

Answer 1

Concepts and reason

The given problem can be solved by using the condition for constructive interference.

To find the thickness of the oil slick, use the condition for constructive interference. Use the expression for the wavelength of ray in air medium and oil medium relation. The expression for the wavelength of ray in air medium and water medium relation can be used to solve the problem.

Fundamentals

The condition for constructive interference can be expressed as,

$2nt = m\lambda$

Here, $n$ is the refractive index, $m$ is the order and $\lambda$ is the wavelength of the radiation.

The light from air-oil interface has a phase reversal of ${180^ \circ }$ . The light from oil-water interface has phase reversal of ${180^ \circ }$ .

The interference between these two rays gives zero or two phases reversal.

For minimum thickness, $m = 1$

The expression for the minimum thickness can be expressed as,

${t_{\min }} = \frac{\lambda }{{2n}}$

The light from the oil water gives no phase reversal. The light from air-oil interface has a phase reversal of ${180^ \circ }$ . The light from oil-water interface has no phase reversal.

The interference between these two rays gives phase reversal. The condition for constructive interference is given as,

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

Here, $n$ is the refractive index, $m$ is the order and $\lambda$ is the wavelength of the radiation.

For minimum thickness, $m = 0$

The expression for minimum thickness can be expressed as,

${t_{\min }} = \frac{\lambda }{{4n}}$

The wavelength of ray in air medium and oil medium are related as follows as,

${n_{{\rm{air}}}}{\lambda _{{\rm{air}}}} = {n_{{\rm{oil}}}}{\lambda _{{\rm{oil}}}}$

Here, ${n_{{\rm{air}}}}$ is the refractive index of air and ${n_{{\rm{oil}}}}$ is the refractive index of oil.

(A)

The minimum thickness of the slick of the oil can be calculated as,

${t_{\min }} = \frac{\lambda }{{2n}}$

Substitute $750 {\rm{nm}}$ for $\lambda$ and $1.2$ for $n$ in the expression for the minimum thickness of the slick of the slick of the oil.

$\begin{array}{c}\\{t_{\min }} = \frac{{\left( {750 {\rm{nm}}} \right)}}{{2\left( {1.2} \right)}}\\\\ = 312.5 {\rm{nm}}\\\end{array}$

(B)

The minimum thickness of the oil slick at the spot can be calculated as,

${t_{\min }} = \frac{\lambda }{{4n}}$

Substitute $750 {\rm{nm}}$ for $\lambda$ and $1.5$ for $n$ in the expression for the minimum thickness of the slick of the slick of the oil.

$\begin{array}{c}\\{t_{\min }} = \frac{{\left( {750 {\rm{nm}}} \right)}}{{4\left( {1.5} \right)}}\\\\ = 125 {\rm{nm}}\\\end{array}$

(C)

The wavelength of light in the air can be calculated as,

${n_{{\rm{air}}}}{\lambda _{{\rm{air}}}} = {n_{{\rm{oil}}}}{\lambda _{{\rm{oil}}}}$

The expression for wavelength of light in oil can be expressed as,

${\lambda _{{\rm{oil}}}} = \frac{{{n_{{\rm{air}}}}{\lambda _{{\rm{air}}}}}}{{{n_{{\rm{oil}}}}}}$

For longest wavelength $k = 1$ ,

$2t = k{\lambda _{{\rm{oil}}}}$

Substitute $\frac{{{n_{{\rm{air}}}}{\lambda _{{\rm{air}}}}}}{{{n_{{\rm{oil}}}}}}$ for ${\lambda _{{\rm{oil}}}}$ in the thickness and wavelength relation.

$2t = k\left( {\frac{{{n_{{\rm{air}}}}{\lambda _{{\rm{air}}}}}}{{{n_{{\rm{oil}}}}}}} \right)$

Rearrange the equation to get the value of ${\lambda _{{\rm{air}}}}$ .

${\lambda _{{\rm{air}}}} = \frac{{2t{n_{{\rm{oil}}}}}}{{{n_{{\rm{air}}}}}}$

Substitute $200 {\rm{nm}}$ for $t$ , $1.5$ for ${n_{{\rm{oil}}}}$ , $1$ for ${n_{{\rm{air}}}}$ in the expression of ${\lambda _{{\rm{air}}}}$ .

$\begin{array}{c}\\{\lambda _{{\rm{air}}}} = \frac{{2\left( {200 {\rm{nm}}} \right)\left( {1.5} \right)}}{{\left( 1 \right)}}\\\\ = 600 {\rm{nm}}\\\end{array}$

The wavelength of light in the water can be calculated as,

${n_{{\rm{air}}}}{\lambda _{{\rm{air}}}} = {n_{{\rm{water}}}}{\lambda _{{\rm{water}}}}$

Rearrange the equation to get the value of wavelength of light in water.

${\lambda _{{\rm{water}}}} = \frac{{{n_{{\rm{air}}}}{\lambda _{{\rm{air}}}}}}{{{n_{{\rm{water}}}}}}$

Substitute for $600 {\rm{nm}}$ ${\lambda _{{\rm{air}}}}$ , $1.33$ for ${n_{{\rm{water}}}}$ , $1$ for ${n_{{\rm{air}}}}$ in the expression of ${\lambda _{{\rm{water}}}}$ .

$\begin{array}{c}\\{\lambda _{{\rm{water}}}} = \frac{{\left( 1 \right)\left( {600 {\rm{nm}}} \right)}}{{\left( {1.33} \right)}}\\\\ = 451.1 {\rm{nm}}\\\end{array}$

Ans: Part A

The minimum thickness of the oil slick at the spot is $312.5 {\rm{nm}}$ .

Part B

The minimum thickness of the oil slick at the spot is $125 {\rm{nm}}$ .

Part C

The longest wavelength of light in water is $451.1 {\rm{nm}}$ .