Light of wavelength 624 nm is incident perpendicularly on a soap film (n = 1.33) suspended in air. What are the (a) least and (b) second least thicknesses of the film for which the reflections from the film undergo fully constructive interference?

Bright colors reflected from thin oil on water and soap bubbles are a consequence of light interference. Due to the constructive interference of light reflected from the front and back surfaces of the thin film, these bright colors can be seen. For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:

$\mathbf{2}\mathbf{L}\mathbf{=}\left(m+\frac{1}{2}\right)\frac{\mathbf{\lambda }}{{\mathbf{n}}_{\mathbf{2}}}\mathbf{m}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{.}\mathbf{..}$(Maxima—bright film in the air)

where λ is the wavelength of the light in air, L is its thickness, and n₂ is the film’s refraction index.

The least and the 2^nd least thickness are associated with $\text{m}=0,1$the order of maximum intensity. The thickness corresponding to this order of maxima is

For $\text{m}=0$;

$\begin{array}{c}{\mathrm{L}}_{\mathrm{o}}=\left(\mathrm{m}+\frac{1}{2}\right)\frac{\mathrm{\lambda }}{2{\mathrm{n}}_2}\\ =\left(0+\frac{1}{2}\right)\frac{624\mathrm{nm}}{2\left(1.33\right)}\\ =117\mathrm{nm}\end{array}$

For $\text{m}=1$;

$\begin{array}{c}{\mathrm{L}}_1=\left(\mathrm{m}+\frac{1}{2}\right)\frac{\mathrm{\lambda }}{2{\mathrm{n}}_2}\\ =\left(1+\frac{1}{2}\right)\frac{624\mathrm{nm}}{2\left(1.33\right)}\\ =352\mathrm{nm}\end{array}$

Thus, the least thickness is 117 nm, and the 2^nd least thickness is 352 nm.