Question

A glass plate of refractive index ${\mu }_1=1.5$ is coated with a thin layer of thickness $t$ and refractive index ${\mu }_2=\mathrm{1.8.}$ Light of wavelength $\lambda$ travelling in air is incident normally on the layer. It is partly reflected at the upper and lower surfaces of the layer, and the two reflected rays interfere. If $\lambda =648\text{nm}$ obtain the least, value of $t$ for which rays interfere constructively.(in $\text{nm}$) . (integers only)

Answer 1

Given, ${\mu }_1=1.5;{\mu }_2=1.8;{\mu }_3=1$



Let, $r_1$ and $r_2$ are the two rays considered for the interference.

The ray $r_1$ gets reflected at the denser medium, hence it suffers phase difference $\varphi =\pi .$ The corresponding path difference is,

$\Delta x_1=\varphi \times \frac{\lambda }{2\pi }=\frac{\lambda }{2}$

The ray $r_2$ gets reflected at the rarer medium, thus it suffers a path difference of

$\Delta x_2=2{\mu }_{2}t$

The net path difference is,

$\Delta x=\Delta x_2-\Delta x_1=2{\mu }_{2}t-\frac{\lambda }{2}$

For the constructive interference,

$\text{Path difference}\left(\Delta x\right)=n\lambda$

$2{\mu }_{2}t-\frac{\lambda }{2}=n\lambda$

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

For $t_{min};(n=0)$

$\therefore t_{min}=\frac{\lambda }{4{\mu }_2}=\frac{648}{4\times 1.8}=90\text{nm}$