A uniform thin film of alcohol (n=1.36) lies on a flat glass plate (n=1.56). When monochromatic...

A uniform thin film of alcohol (n=1.36) lies on a flat glass plate (n=1.56). When monochromatic light, whose wavelength can be changed, is incident normally, the reflected light is a minimum for ?=515nm and a maximum for ?=644nm.

What is the minimum thickness of the film?

tmin = .....?..... nm

Solutions

Expert Solution

When the incident llight reflects from top of the alcohol has a phase change of $\phi$ 1 = $\Pi$

The wave reflects from the glass at the bottom of the alcohol has a phase change is

$\phi$ 2 = (2t/ $\lambda$ _film) *2 $\Pi$ + $\Pi$

For constructive interference the net phase change is

$\phi$ = $\phi$ 2 - $\phi$ 1 = m1 *2 $\Pi$ , m1 = 1,2.3.....

then t =1/2* $\lambda$ _film *m1 = 1/2*( $\lambda$ 1/n_film)*m1

For destructive interference the net phase change is

$\phi$ = $\phi$ 2 - $\phi$ 1 = 1/4* $\lambda$ _film *(2*m2+1) = 1/2*( $\lambda$ 2/n_film)*(2*m2+1)

from the two equations

1/2*( $\lambda$ 1/n_film)*m1 = 1/2*( $\lambda$ 2/n_film)*(2*m2+1)

(2m2+1)/m1 = $\lambda$ 1 / $\lambda$ 2 = 644/515 = 1.25

Thus we have

m1 = m2 = 2

then thickness of the film

t = 1/2*(644/1.36)*2

t = 473.52 nm

genius_generous answered 1 year ago

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