Question

A glass surface is coated by an oil film of uniform thickness $1.00\times 10^{- 4}cm.$ The index of refraction of the oil is $1.25$ and that of the glass is $1.50$ . Some of the wavelengths in visible region $\left(400 nm - 490 nm\right)$ are completely transmitted by the oil film under normal incidence. One of the wavelength transmitted completely in visible region is $\frac{n \times 10^{- 6}}{11}m.$ Find the value of $n$ .

Answer 1

Optical path difference for the light transmitted through oil is
$\Delta x=2\mu lcosr$
For normal incidence

But the interface between oil and glass will produce an additional path difference of $\frac{\lambda }{2}$ . Therefore, effective path difference
$\Delta x=2\mu t+\frac{\lambda }{2}$
For constructive interference in transmitted light
$2\mu t+\frac{\lambda }{2}=n\lambda ,n=1,2,\ldots \ldots $
or $2\mu t=\left(2 n - 1\right)\frac{\lambda }{2}$
or $\lambda =\frac{4 \mu t}{\left(2 n - 1\right)}=\frac{4 \times 1 . 25 \times 1 \times \left(10\right)^{- 6}}{2 n - 1}=\frac{5 \times \left(10\right)^{6}}{\left(2 n - 1\right)}$
For $n=1,\lambda =5\times 10^{- 6}m=5000nm$
For $n=2,\lambda =\frac{5 \times 10^{- 6}}{3}m=1666.67nm$
For $n=3,\lambda =\frac{5 \times 10^{- 6}}{5}m=1000nm$
For $n=4,\lambda =\frac{5 \times 10^{- 6}}{7}m=714.29nm$
For $n=5,\lambda =\frac{5 \times 10^{- 6}}{5}m=555.55nm$
For $n=6,\lambda =\frac{5 \times 10^{- 6}}{11}m=454.54nm$
The wavelength which are strongly transmitted in visible range are $:714.29nm$ , $555.55nm$ and $454.54nm$ .