What are the three smallest non-zero thicknesses of soapy water \((n = 1.33)\) on Plexiglas if it appears green (constructively reflecting \(520 - nm\) light) when illuminated perpendicularly by white light? Explicitly show how you follow the steps in Problem Solving Strategies for Wave Optics.

When the wavelength of a wave is measured, the distance between the same locations between two succeeding waves is calculated.

Wavelength of the red soapy water is:

\(\begin{aligned}\lambda &= 520{\rm{ }}nm\\ &= \dfrac{{520}}{{{{10}^9}}}\\ &= 520 \times {10^{ - 9}}{\rm{ }}m\end{aligned}\).

Value for index of redfraction of soapy water is: \(n = 1.33\).

To find required minimum thickness, use relation for constructive interference of reflected light –

\(2nd = 3\lambda \)

Afterward simplifying above relation, the relation for finding minimum thickness is obtained –

\(\begin{aligned}\dfrac{{2nd}}{2} &= \dfrac{3}{2}\lambda \\nd &= \dfrac{{3\lambda }}{2}\\\dfrac{{nd}}{n} &= \dfrac{{3\lambda }}{{2n}}\\d &= \dfrac{{3\lambda }}{{2n}}{\rm{ }}Equation(1)\end{aligned}\)

Now, plug in values in equation\((1)\) for minimum thickness and solve this equation –

\(\begin{aligned}d &= \dfrac{{3\lambda }}{{2n}}\\d &= \dfrac{{3 \cdot 520 \cdot {{10}^{ - 9}}{\rm{ }}m}}{{2 \cdot 1.33}}\\d &= \dfrac{{1560 \cdot {{10}^{ - 9}}{\rm{ }}m}}{{2.66}}\\d &= 586 \cdot {10^{ - 9}}{\rm{ }}m{\rm{ }}(Yellow{\rm{ }}colour)\end{aligned}\)

Therefore, the value for thickness is obtained as \(d = 586 \cdot {10^{ - 9}}{\rm{ }}m\)(yellow colour).