Question

An oil slick on water is \({\rm{120 nm}}\) thick and illuminated by white light incident perpendicular to its surface. What colour does the oil appear (what is the most constructively reflected wavelength), given its index of redfraction is \({\rm{1}}{\rm{.40}}\)?

Short answer

The wavelength and colour of oil that is most constructively reflected, is \(\lambda = 672 \cdot {10^{ - 9}}{\rm{ m}}\) and red colour respectively.

Step-by-step solution

Concept Introduction

The distance between the identical places between two subsequent waves is determined when the wavelength of a wave is measured.

Information Provided

Thickness of oil slick on water is:

\(\begin{aligned}d &= 120{\rm{ nm}}\\ &= \dfrac{{120}}{{{{10}^9}}}\,{\rm{m}}\\ &= 120 \times {10^{ - 9}}{\rm{ m}}\end{aligned}\).

Value for index of redfraction of oil is: \(n = 1.4\).

Calculation for wavelength

To find required wavelength and colour, use relation for constructive interference of reflected light –

\(2nd = \dfrac{1}{2}\lambda \)

Afterward simplifying above relation, the relation for finding wavelength is obtained –

\(\begin{aligned}2nd &= \dfrac{1}{2}\lambda \\2 \cdot \left( {2nd} \right) &= \dfrac{1}{2}\lambda \cdot 2\\4nd &= \lambda {\rm{ }}................{\rm{ (1)}}\end{aligned}\)

Now, plug in values in equation\({\rm{(1)}}\)for wavelength and solve this equation –

\(\begin{aligned}\lambda &= 4 \cdot n \cdot d\\\lambda &= 4 \cdot \left( {1.4} \right) \cdot \left( {120 \cdot {{10}^{ - 9}}{\rm{ m}}} \right)\\\lambda &= 672 \cdot {10^{ - 9}}{\rm{ m }}\left( {{\rm{Red Colour}}} \right)\end{aligned}\)

Therefore, the value for wavelength is obtained as \(\lambda = 672 \cdot {10^{ - 9}}{\rm{ m}}\) (red colour).