Question

Monochromatic light of wavelength $640\text{nm}$ falls normally on a glass plate of refractive index $\mathrm{1.6.}$ The waves reflected from the upper and lower faces of the film will interfere constructively if the least thickness of the plate is .

• A. $100\text{nm}$
• B. $200\text{nm}$
• C. $300\text{nm}$
• D. $400\text{nm}$

Answer 1

The correct option is A $100\text{nm}$

When light falls on a surface, it is partly reflected and partly refracted. Thus light from source $S$ is reflected as wave $1$ at $A$ of the top surface of the plate, refracted into the plate and reflected as wave $2$ at $B$ of the bottom surface. On reaching $A,$ the wave $2$ has traversed an optical path length $=\mu t+\mu t=2\mu t$


Furthermore, on reflection from the denser medium, a wave suffers a phase change of $\pi$ which implies a path change of $\lambda /2.$ Hence, on reaching $A$ waves $1$ and $2$ will have a path change of
$\Delta x=2\mu t-\frac{\lambda }{2}$

For constructive interference. $\Delta x=n\lambda$
Hence,
$2\mu t-\frac{\lambda }{2}=n\lambda$

$\Rightarrow 2\mu t=(n+\frac{1}{2})\lambda$

or $t=\frac{(n+\frac{1}{2})\lambda }{2\mu }$
The minimum value of $t$ corresponds to $n=0$, hence

$t_{min}=\frac{\lambda }{4\mu }=\frac{640\text{nm}}{4\times 1.6}=100\text{nm}$