91Ó°ÊÓ

• The refractive index of first medium is $n_1=1.68$.
• The refractive index of the thin film is$n_2=1.59$
• The refractive index of the third medium $n_3=1.50$
• The minimum intensity occurs at wavelength $\mathrm{λ}=342nm$.

Bright colours 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 colours can be seen. For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:

$2L=\left(m+\frac{1}{2}\right)\frac{\mathrm{λ}}{n_2}m=0,1,2,...(\mathrm{Maxima}—\mathrm{bright}\mathrm{film}\mathrm{in}\mathrm{the}\mathrm{air})$

where $\mathrm{λ}$is the wavelength of the light in air, $L$ is its thickness, and $n_2$ is the film’s refractive index.

Here, in this case, light travels in medium with $n_1=1.68$and incident on the film whose refractive index is $n_2=1.59$And then, the light gets reflected of the third surface $n_3=1.50$while travelling through the film. A minimum intensity is observed at $\mathrm{λ}=342nm$and thickness of the film is to be determined at 2nd order of intensity which $m=1$in this case. As a result, the condition for destructive interference or minimum intensity is

$2L=\left(m+\frac{1}{2}\right)\frac{{\mathrm{λ}}_{342}}{n_2}$ (Destructive)

The 2^nd least thickness is

$\begin{array}{rcl}2L& =& \left(1+\frac{1}{2}\right)\frac{342\mathrm{nm}}{1.59}\\ L& =& \frac{3\left(342\mathrm{nm}\right)}{4\left(1.59\right)}\\ & =& 161\mathrm{nm}\end{array}$

Hence the thickness of the thin layer is $161\mathrm{nm}$.