Question

A plane wave of monochromatic light is incident normally on a uniform thin film of oil that covers a glass plate. The wavelength of the source can be varied continuously. Fully destructive interference of the reflected light is observed for wavelengths of $\mathbf{500}\mathbf{}\mathbf{\text{nm}}$and $\mathbf{700}\mathbf{}\mathbf{\text{nm}}$and for no wavelengths in between. If the index of refraction of the oil is $\mathbf{1}\mathbf{.}\mathbf{30}$and that of the glass is $\mathbf{1}\mathbf{.}\mathbf{50}$, find the thickness of the oil film.

Short answer

The thickness of the oil film is $673\text{nm}$.

Step-by-step solution

Given Data.

• The refractive index of the oil film is$n_2=1.30$
• The refractive index of the glass is$n_3=1.50$
• The minimum intensity is observed at wavelengths role="math" localid="1663025634127" ${\lambda }_{\min 1}=500\text{nm}$androle="math" localid="1663025642919" ${\lambda }_{\min 2}=700\text{nm}$

Interference in thin films

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:

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

where$\lambda$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 an air medium and incident on the oil film whose refractive index is higher than air. And then, the light gets reflected of the thick glass while traveling through oil, and again the refractive index is higher than acetone. As a result, the condition for destructive interference is

$2L=\left(m+\frac{1}{2}\right)\frac{{\lambda }_{600}}{n_2}$ (Destructive)

Determining thickness of the oil film.

For destructive interference of $500\text{nm}$reflected light, the value of thickness for the first few orders is

$\begin{array}{l}m=0;L=\left(0+\frac{1}{2}\right)\frac{500nm}{2\left(1.30\right)}=96\text{nm}\\ m=1;L=\left(1+\frac{1}{2}\right)\frac{500nm}{2\left(1.30\right)}=289\text{nm}\\ m=2;L=\left(2+\frac{1}{2}\right)\frac{500nm}{2\left(1.30\right)}=481\text{nm}\\ m=3;L=\left(3+\frac{1}{2}\right)\frac{500nm}{2\left(1.30\right)}=673\text{nm}\end{array}$

For destructive interference of $700\text{nm}$reflected light, the value of thickness for the first few orders is

role="math" localid="1663026287442" $\begin{array}{l}m=0;L=\left(0+\frac{1}{2}\right)\frac{700nm}{2\left(1.30\right)}=135\text{nm}\\ m=1;L=\left(1+\frac{1}{2}\right)\frac{700nm}{2\left(1.30\right)}=404\text{nm}\\ m=2;L=\left(2+\frac{1}{2}\right)\frac{700nm}{2\left(1.30\right)}=481\text{nm}\\ m=3;L=\left(3+\frac{1}{2}\right)\frac{500nm}{2\left(1.30\right)}=673\text{nm}\end{array}$

The least common thickness for$500\text{nm}$ and $700\text{nm}$reflected light is $673\text{nm}$. Hence the thickness of the oil film is $673\text{nm}$.