Question

g Visible light travels in air, which has an index of refraction of 1.0. It is incident on plastic which is covered by a coating (such that light hits the coating before the plastic). The coating has an index of refraction of 1.6 and is 0.50 microns thick. The ray reflected off the air-coating interface and the ray reflected off the coating-plastic interface experience interference. What frequencies could the light be

Answer 1

Answer:

The interference is constructive. The plastic used has an index of refraction of 1.7

Explanation:

Given that :

Visible light travels in air, which has an index of refraction of 1.0

i.e $\mathbf{n_{air} = 1.0}$

The coating has an index of refraction of 1.6

i.e $\mathbf{n_{coat} = 1.6}$

Since the index of refraction of air is incident on the plastic which is covered by a coating ; then $\mathbf{n_{plastic} = 1.7}$

Now; for constructive interference:

$\mathbf{2 nt_{coat} =m \lambda _ m}$

$\mathbf{\lambda _ m =\frac{2 nt_{coat}}{m} }$

where ;

m =1, 2, ... n

For m = 1

$\mathbf{\lambda _ 1 =\frac{2 *1.6*0.50*10^{-6}}{1} }$

$\mathbf{\lambda _ 1 =1.6*10^{-6} \ m}$

$\mathbf{f_1 =\frac{c}{\lambda_1} }$

$\mathbf{f_1} =\frac{3*10^8}{1.6*10^{-6} }$

$\mathbf{f_1 =1.875*10^{14} \ Hz}$

For m =2

$\mathbf{\lambda _ 2 =\frac{2 *1.6*0.50*10^{-6}}{2} }$

$\mathbf{\lambda _ 1 =8.0*10^{-7}m}$

$\mathbf{f_2 =\frac{c}{\lambda_2} }$

$\mathbf{f_2} =\frac{3*10^8}{8*10^{-7} }$

$\mathbf{f_2 =3.75*10^{14} \ Hz}$

$\mathbf{f_n=n(1.875)*10^{14} \ Hz}$     for n = 1, 2 , .......