Question

A plastic film with index of refraction 1.85 is put on the surface of a car window to increase the reflectivity and thus to keep the interior of the car cooler. The window glass has index of refraction 1.52.

Part A

What minimum thickness is required if light with wavelength 550 nm in air reflected from the two sides of the film is to interfere constructively?

Part B

It is found to be difficult to manufacture and install coatings as thin as calculated in part (a). What is the next greatest thickness for which there will also be constructive interference?

Answer 1

\(\underline{\text { STEP-II: }}\) For the constructive interference for reflected rays Path difference \(=\mathrm{m} \lambda\) \(2 \mu \mathrm{t}-\frac{\lambda}{2}=\mathrm{m} \lambda\)

That is \(2 \mu t=(2 m+1) \frac{\lambda}{2} \ldots \ldots \ldots . .(1)\)

Where \(t\) is the thickness of the film

\(\underline{\text { STEP-III: }}\)

(a) For minimum thickness \(m=0\) in equation (1) That is \(\begin{aligned} 2 \mu t &=\frac{\lambda}{2} \\ t &=\frac{\lambda}{4 \mu} \\ &=\frac{550 \times 10^{-9} \mathrm{~m}}{4 \times 1.85} \\ &=74.3 \mathrm{~nm} \end{aligned}\)

(b) For the next greatest thickness \(\mathrm{m}=1\) in equation (1)

$$ \text { That is } \begin{aligned} 2 \mu t &=\frac{3 \lambda}{2} \\ t &=\frac{3 \lambda}{4 \mu} \\ &=\frac{3 \times 550 \times 10^{-9} \mathrm{~m}}{4 \times 1.85} \\ &=223 \mathrm{~nm} \end{aligned} $$