Question

78. A film of oil on water will appear dark when it is very thin, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the oil can be and appear dark at all visible wavelengths? Oil has an index of refraction of 1.40.

Answer 1

To solve this problem it is necessary to apply the concepts related to the condition of path difference for destructive interference between the two reflected waves from the top and bottom of a surface.

Mathematically this expression can be described under the equation

$\delta = 2nt$

Where

n = Refractive index

t = Thickness

In terms of the wavelength the path difference of the reflected waves can be described as

$\delta = \frac{\lambda}{4}$

Where

\lambda = Wavelenght

Equation the two equations we have that

$2nt = \frac{\lambda}{4}$

$t = \frac{\lambda}{8n}$

Our values are given as

$\lambda = 380nm \rightarrow$ Wavelength of light

$n = 1.4$

$t = \frac{380nm}{8*1.4}$

$t = 33.93nm$

Therefore the minimum thickness of the oil for destructive interference to occur is approximately 34.0 nm