Question

A thin film of alcohol (n = 1.36) lies on a flat glass plate (n = 1.51). When monochromatic light, whose wavelength λ can be changed, is incident normally, the reflected light is minimum for wavelength λ = 512 nm and a maximum for wavelength = 640 nm. What is the minimum thickness of the film?

Answer 1

Answer:

The thickness of the film, t = 470.59 nm

Explanation:

$\lambda_{film} = \frac{\lambda}{n_{film} }$

For constructive interference, the net phase change, $\phi = \phi_{2} - \phi_{1} = 2\pi m_{1}$

Then the thickness,  $t = 0.5 \lambda_{film} m_{1}$

m₁ = 1,2,3........

$t = 0.5\frac{\lambda_{1} }{n_{film} } m_{1}$..............(i)

For destructive interference, the net phase change is  $\phi = \phi_{2} - \phi_{1} = 2(m_{2} + 1)\pi$

m₂ = 1,2,3........

thickness, $t = 0.5\frac{\lambda_{1} }{n_{film} }(2 m_{2}+1)$..............(ii)

Equating (i) and (ii)

$0.5\frac{\lambda_{1} }{n_{film} } m_{1} = 0.5\frac{\lambda_{1} }{n_{film} }(2 m_{2}+1)$

$\frac{2m_{2} + 1}{2m_{1} } = \frac{\lambda_{1} }{\lambda_{2}} = \frac{640}{512}$

$\frac{2m_{2} + 1}{2m_{1} } = 1.25$.............(iii)

From (iii), When m₁ =2, m₂ = 2

The thickness , $t = 0.5\frac{\lambda_{1} }{n_{film} } m_{1}$ then becomes

$t = 0.5\frac{640}{1.36} *2$

t = 470.59 nm