91Ó°ÊÓ

• The refractive index of first medium is $n_1=1.60$.
• The refractive index of the thin film is$n_2=1.40$
• The refractive index of the third medium$n_3=1.80$
• The thickness of the layer is $L=200nm$.

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:

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

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

Here, in this case, light travels in medium with$n_1=1.60$ and incident on the film whose refractive index is$n_2=1.40$ and the thickness of the layer is$L=200nm$ the reflected light has no phase change. And then, the reftracted light gets reflected of the third surface$n_3=1.80$ while travelling through the film. This results in$180°$ phase change as the light is reflecting of a denser medium. As a result, the condition for destructive interference or minimum intensity is

$2L=\frac{m\mathrm{λ}}{n_2}$ (destructive)

The minimum intensity is observed for wavelength,

role="math" localid="1663143094524" $m=1;\mathrm{λ}=\frac{2L{n}_2}{m}=\frac{2\left(200\mathrm{nm}\right)\left(1.40\right)}{1}=560\mathrm{nm}$

As $560\mathrm{nm}$is in the visible range, hence the wavelength of the reflected light be $560\mathrm{nm}$.