91Ó°ÊÓ

Light interference is the cause of the vivid colors that thin oil on water and soap bubbles reflect. These vivid hues can be seen because of the light reflection from the thin film's front and rear surfaces' constructive interference.

For a perpendicular incident beam, the maximum intensity of light from the thin film satisfies the condition:

$2\mathrm{L}=\left(\mathrm{m}+\frac{1}{2}\right)\frac{\mathrm{λ}}{{\mathrm{n}}_2}\text{ â¶Ä‰â¶Ä‰}\mathrm{m}=0,\text{ }1,\text{ }2,\mathrm{...}$(Maxima—bright film in the air)

$2\mathrm{L}=\mathrm{m}\frac{\mathrm{λ}}{{\mathrm{n}}_2}\text{ â¶Ä‰â¶Ä‰}\mathrm{m}=1,\text{ }2,\text{ }3,\mathrm{..}$ (Minima)

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 a medium with $n_1=1.40$ and incident on the thin layer whose refractive index is $n_2=1.46$ and the reflected light has $180°$ phase change as the light is reflected off the denser medium. And then, the refracted light gets reflected of the back surface $n_3=1.75$ while traveling through the film. This results in $180°$ phase change. The total phase difference between $r_1$ and $r_2$ is still zero. As a result, the condition for destructive interference or minimum intensity is

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

The 2nd least $(m=1)$ thickness of the thin layer is

$\begin{array}{c}L=\left(1+\frac{1}{2}\right)\frac{482\text{ }nm}{2\left(1.46\right)}\\ =248\text{ }nm\end{array}$

Hence, the 2nd least thickness of the thin layer is$248\text{ }nm$ .