A glass plate of reflective index ${\mu }_3=1.5$ is coated with a thin layer of thickness t and refractive index ${\mu }_2=1.8.$ Light of wavelength $\lambda$ travelling in air is incident normally on the layer. It is partly reflected at the upper and the lower surfaces of the layer and the two reflected rays interfere. If $\lambda =648nm$ The least value of t for which the waves interfere constructively is

Refer to the following figure. a ray of light travelling in air $({\mu }_1=1)$ falls normally on a thin layer $({\mu }_2=1.8)$ of thickness t. It is partly reflected at point P as wave 1 and partly reflected as wave 2 on meeting the surface of the glass plate $({\mu }_3=1.5)$ is reflected at point Q and travels along QP.

${\Delta }_2=$Refractive index of layer $x2\left(PQ\right)={\mu }_2\times 2t=2{\mu }_{2}t$

Optical path difference between waves 1 and 2 at point p is $\Delta ={\Delta }_2-{\Delta }_1=2{\mu }_{2}t-\frac{\lambda }{2}$

Now for constructive interference, $\Delta =n\lambda ,n=0,1,\mathrm{2.....}$ Or $2{\mu }_{2}t-\frac{\lambda }{2}=n\lambda$ or $2{\mu }_{2}t=(n+\frac{1}{2})\lambda$ Or $t=\frac{(n+\frac{1}{2})\lambda }{2{\mu }_2}$

The minimum value of t corresponds to $n=0.$ Hence $t_{\min }=\frac{\lambda }{4{\lambda }_2}=\frac{648nm}{4\times 1.8}=90nm$