Chapter 23, Problem 100QAP

To determine

(a)

The minimum thickness of coating needed to cancel visible light of wavelength $400nm$ in the top of the coating in air.

Answer to Problem 100QAP

The intensity of polarized light that passes through the stack of polarizing sheets is $9.38W/m^2.$

Explanation of Solution

Given:

Refractive index for coating = $n_{coat}=1.45$

Refractive index for solar cell panel = $n_{cell}=3.5$

Wavelength of light = $\lambda =400nm$

Formula used:

Formula for destructive interference due to thin film,

  $\begin{array}{l}2nt=\left(m+\frac{1}{2}\right)\lambda \\ t=\text{ thickness of film}\\ \text{n = refractive index for film}\\ \text{m = number of order of fringes}\\ \lambda \text{=wavelength of light}\end{array}$

Calculation:

Since, path of light wave on the top and bottom face of the film changes by $\frac{\lambda }{2}$.

So, reflected waves would be in phase. So, path difference between top and bottom surface of the film would be as,

  $\Delta x=2n_{coat}t=\left(m+\frac{1}{2}\right)\lambda (\text{for destructive interference)}$

  $\begin{array}{l}t=\frac{\left(m+\frac{1}{2}\right)\lambda }{2n_{coat}}\\ \text{for minimum thickness, m = 0, then,}\\ {\text{t}}_{\min }=\frac{\left(0+\frac{1}{2}\right)400nm}{2(1.45)}\\ t_{\min }=69nm\end{array}$

Conclusion:

Thus, the minimum thickness of coating that cancel visible light of wavelength $400nm$ is $69nm$.

To determine

(b)

The wavelength of visible light that cancel and reinforce the reflected light when thickness is increases by three times to its initial value

Answer to Problem 100QAP

The wavelength of the reflected light that cancels in air is $400nm$ and the wavelength of light that reinforces is $600nm.$

Explanation of Solution

Given:

Thickness of coating = $t\text{'}=3t=(3)(69nm)=207nm$

Refractive index for coating = $n_{coat}=1.45$

Refractive index for solar cell panel = $n_{cell}=3.5$

Formula used:

Formula for destructive interference due to thin film,

  $2nt=\left(m+\frac{1}{2}\right)\lambda$

  $\begin{array}{l}\text{for constructive interference,}\\ 2nt=m\lambda \\ t=\text{ thickness of film}\\ \text{n = refractive index for film}\\ \text{m = number of order of fringes}\\ \lambda \text{=wavelength of light}\end{array}$

Calculation:

For destructive interference,

  $\begin{array}{l}2nt\text{'}=\left(m+\frac{1}{2}\right)\lambda \\ \lambda =\frac{2nt\text{'}}{m+\frac{1}{2}}=\frac{2\times 207nm\times 1.45}{m+\frac{1}{2}}=\frac{600nm}{m+\frac{1}{2}}\\ \text{when m}\text{= 0,}\\ \lambda =1200nm(Thisdoes\text{ not comes under visible light)}\\ \text{When m = 1,}\\ \lambda =400nm(\text{This comes in visible region}\text{.)}\\ \text{when m = 2,}\\ \lambda =240nm(This\text{ is not in visible region}\text{.)}\end{array}$

As we increase m, wavelength would be less than 240nm.

So, there is only one reflected wavelength lies in visible region that cancels is $\lambda =400nm$

Now, for constructive interference,

  $\begin{array}{l}2nt\text{'}=m\lambda \\ \lambda =\frac{2nt\text{'}}{m}=\frac{2(207nm)(1.45)}{m}=\frac{600nm}{m}\\ \text{when m = 1,}\\ \lambda =600nm\text{(This is in visible region}\text{.)}\\ \text{for m = 2,3,4 and so on, wavelength of light does not fall in }\\ \text{visible region}\text{. }\\ \text{So, only 600 nm reinforce for the reflected light that lies }\\ \text{in visible region}\text{.}\end{array}$

Conclusion:

Thus, the wavelength of reflected light that cancels is $400nm$ and the wavelength of light that reinforce the reflected light is $600nm.$