Example 14.2: Interference in thin films

Question: A soap bubble 250 nm thick is illuminated by white light. The index of refraction of the soap film is $1.36$. Which colours are not seen in the reflected light? Which colours appear strong in the reflected light? What colour does the soap film appear at normal incidence?
 
Solution: For destructive interference, we must have $n\,t=m\,\lambda/2$. Thus, the wavelengths that are not reflected satisfy

$$\lambda_m = \frac{2\,n\,t}{m},$$

where $m=1,2,3,\cdots$. It follows that

$$\lambda_1 = \frac{(2)\,(1.36)\,(250\times 10^{-9})}{(1)} = 680\,{\rm nm},$$

and

$$\lambda_2 = \frac{(2)\,(1.36)\,(250\times 10^{-9})}{(2)} = 340\,{\rm nm}.$$

These are the only wavelengths close to the visible region of the electromagnetic spectrum for which destructive interference occurs. In fact, 680 nm lies right in the middle of the red region of the spectrum, whilst 340 nm lies in the ultraviolet region (and is, therefore, invisible to the human eye). It follows that the only non-reflected colour is red.

For constructive interference, we must have $n\,t=(m+1/2)\,\lambda/2$. Thus, the wavelengths that are strongly reflected satisfy

$$\lambda_m' = \frac{2\,n\,t}{m+1/2},$$

where $m=0,1,2,\cdots$. It follows that

$$\lambda_1' = \frac{(2)\,(1.36)\,(250\times 10^{-9})}{(1/2)} = 1360\,{\rm nm},$$

and

$$\lambda_2' = \frac{(2)\,(1.36)\,(250\times 10^{-9})}{(3/2)} = 453\,{\rm nm},$$

and

$$\lambda_3' = \frac{(2)\,(1.36)\,(250\times 10^{-9})}{(5/2)} = 272\,{\rm nm}.$$

A wavelength of 272 nm lies in the ultraviolet region whereas 1360 nm lies in the infrared. Clearly, both wavelengths correspond to light which is invisible to the human eye. The only strong reflection occurs at 453 nm, which corresponds to the blue-violet region of the spectrum.

The reflected light is weak in the red region of the spectrum and strong in the blue-violet region. The soap film will, therefore, possess a pronounced blue colour.