Chapter 19, Problem 63A

To determine

The color(s) of light for the occurrence of complete destructive interference while travelling through a thin film.

Answer to Problem 63A

The colors that undergo destructive interference are those other than the colors in reddish-orange band.

Explanation of Solution

Given:

A glass lens has antireflective coating with a refractive index n=1.2 with a thickness d=125 nm.

Formula Used:

The given situation is that of thin film interference. Here, the thin film is the antireflective coating of the glass lens and during reflection from it some of the colors undergo a constructive interference and some undergo destructive interference.

And, regarding the phase of the reflected waves, the light first enters to the denser medium, thin film from air which is a rarer medium. This indicates a phase inversion on first reflection. Now, the light further travels to the glass medium which is more denser with a refractive index of 1.5 compared to the thin antireflective coating,and hence there is no phase inversion in the second reflection as well

The condition for the type of interference is dependent on the thickness of the film and the wavelength of the light in the film.

If the thickness of the film is $d=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{film}}{2}$ , then those wavelengths will undergo constructive interference.

All other wavelengths will undergo a destructive interference.

The wavelength of the light in film can be expressed in terms of the wavelength in vacuum and the refractive index of the thin film as,

  ${\lambda }_{film}=\frac{{\lambda }_{vacuum}}{n}$

Calculation:

Here the refractive index of the thin film is n=1.2 and the thickness is 125 nm. Therefore, considering the first case of m=0, the wavelength of light in vacuum is given by,

  $\begin{array}{c}{\lambda }_{vacuum}=4nd\\ =4\times 1.2\times 125\times {10}^{-9}\\ =600\text{ nm}\end{array}$

This wavelength falls in the reddish-orange band. Now, considering the value of m=1, the wavelength is given by,

  $\begin{array}{c}{\lambda }_{vacuum}=\left(\frac{2}{3}\right)\times 2nd\\ =\left(\frac{2}{3}\right)\times 2\times 1.2\times 125\times {10}^{-9}\\ =200\text{ nm}\end{array}$

This wavelength does not fall under the visible light. It can be understood that for values of m greater than unity, the wavelength will be less than the visible range.

This implies that, all the wavelengths except those in the reddish-orange band undergoes destructive interference.

Conclusion:

The colors that will undergo destructive interference are all the colors except the ones that come in reddish-orange band.