Interference by Thin Film

INTERFERENCE BY THIN FILM

A ray of light incident on a thin film of thickness 't' gets partially reflected and refracted at A at surface I and thereafter it gets reflected and refracted at B of surface II. The rays after emerging in the first medium interfere. Now the inference will depend upon the path difference between AD and ABC, as beyond CD path difference is zero.

$\Delta$ x = $\mu$ (AB + BC) – AD

= $\mu$ (AE + EB + BL) – AD

= $\mu$ AE–AD+$\mu$(EB+BC) (I)

In $\Delta$ LBF = BC = BF (II)

$\Rightarrow$ $\Delta$ x = $\mu$ AE – AD + $\mu$ EF (III)

In $\Delta$ ECF EF = CP cos r = 2 t cos r (IV)

In $\Delta$ ADC and $\Delta$ AEC

$\frac{AD}{AC}=\sin iand\frac{AE}{AC}=\sin r$

$\Rightarrow$ $\frac{\sin i}{\sin r}=\frac{AD}{AE}=\mu$

$\Rightarrow$ $\mu$ AE = AD (V)

Putting (IV) and (V) in (III) we obtain

$\Delta$ x = 2 $\mu$t cosr

IAD as In reflected at a denser medium it suffers an additional path difference $\lambda$/2

Total path difference the taken place is

2 $\mu$ t cosr – $\lambda /2$

For constructive interference

2 $\mu$ t cosr – $\lambda /2$ = $\mu$ n.

$\Rightarrow$ 2 $\mu$ t cosr = n$\lambda$ + $\lambda$/2 = (2n + 1)$\lambda$/2 maxima

For normal incidence r = 0 $\Rightarrow$ 2 $\mu$ t = (2n + 1) $\lambda /2$ = $\mu$ n

2 $\mu$ t cosr – $\lambda /2$ = (2n – 1) $\lambda$ / 2 $\Rightarrow$ 2 $\mu$ t cosr = n$\lambda$

For normal incidence 2 $\mu$ t = n $\lambda$

Illustration 1: White light is incident normally on a glass plate of thickness 0.50 x 10^-6 m and index of refraction 1.50. Which wavelengths in the visible region (400 nm - 700 nm) are strongly reflected by the plate?

Solution : The light of wavelength $\lambda$ is strongly reflected if the light rays reflected are interfering constructively.

As we know the condition for constructive interference

2$\mu$t = $\left(n+\frac{1}{2}\right)$$\lambda$. Here 2$\mu$t = 2 x 1.5 x (0.5 x 10^-6)m = 1.5 x 10^-6 m.

Putting $\lambda$ = 400 nm, 1.5 x 10^-6 = $\left(n+\frac{1}{2}\right)$400 x 10^-9 $\Rightarrow$ n = 3.25

Similarly, by putting $\lambda$ = 700 nm.

1.5 x 10^-6 = $\left(n+\frac{1}{2}\right)$(700 x 10^-9)

$\Rightarrow$ n = 1.66

Thus, within 400 nm to 700 nm, for integral values of n = 2 and 3.

Now, $\lambda =\frac{4\mu t}{2n+1}=600nm\text{ and 429 nm are strongly reflected}\text{.}$