A thin film of soap solution $\left({\mu }_s=1.4\right)$ lies on the top of a glass plate $\left({\mu }_g=1.5\right)$. When visible light is incident almost normal to the plate, two wavelengths $420 \mathrm{nm}$ and $630 \mathrm{nm}$ are strongly reflected. What is the minimum thickness (in $\mathrm{nm}$) of the soap film?

Solution

For reflection at the air - soap solution interface, the phase difference is $\pi$



For reflection at the interface of soap solution to glass also, there will be a phase difference of $\pi$

$\therefore$ The condition for the maximum intensity

$= 2\mu t=n\lambda$

For n, $n{\lambda }_1=\left(n-1\right) {\lambda }_2$

$n\times 420=\left(n-1\right)630$

$\therefore$ $n\left(630-420\right)=630$

$\Rightarrow$ $n\left(210\right)=630$

$\Rightarrow$ $n=\frac{630}{210}$

$\Rightarrow$ $n=3$

This is the maximum order, where they coincide

$2\times 1.4\times t=3\times 420$

$\Rightarrow$ $t=\frac{3\times 420}{2\times 1.40}=450\mathrm{ }\mathrm{nm}$

Hence, the correct answer is 450.