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 wavelength in visible region (400nm – 490nm) 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.

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

$r=0,\cos r\to 1\Delta x=2\mu t$
But the interface between oil and glass will produce an additional path difference of  $\frac{\lambda }{2}$. There fore effective path difference $\Delta x=2\mu t+\frac{\lambda }{2}$
For constructive interference in transmitted light $2\mu t+\frac{\lambda }{2}=n\lambda ,n=1,2,\mathrm{............}$   Or

$2\mu t=(2n-1)\frac{\lambda }{2}$  or  $\lambda =\frac{4\mu t}{(2n-1)}=\frac{4\times 1.25\times 1\times {10}^{-6}}{2n-1}=\frac{5\times {10}^6}{(2n-1)}$
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,  $\lambda =\frac{5\times {10}^{-6}}{5}m=1000$nm
For n = 4,  $\lambda =\frac{5\times {10}^{-6}}{7}m=714.29$nm
For n = 5,  $\lambda =\frac{5\times {10}^{-6}}{5}m=555.55$nm
For n = 6,  $\lambda =\frac{5\times {10}^{-6}}{11}m=\mathrm{4.54.54}$nm
The wavelength which are strongly transmitted visible range are : 714.29 nm, 555.55 nm and 454.54 nm.