Question

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 incident light is almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths $420nm$ and $630nm$. The minimum thickness of the soap solution is

• A. 420 nm
• B. 450 nm
• C. 630 nm
• D. 1260nm

Answer 1

Answer: (B)

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

For reflection at the interface of soap solution to the glass also there will be a phase difference of $\pi$.
$\therefore$ The condition for maximum intensity $=2\mu t=n\lambda ...(1)$
For $n,n{\lambda }_1=(n-1){\lambda }_2$
$n(420)=(n-1)630$
$n(630-420)=630$
$\therefore n(210)=630$
$\Rightarrow n=3$
This is the maximum order where they coincide,
From Eq. (i),
$\therefore 2\times 1.4\times t=3\times 420$
$\Rightarrow t=\frac{2\times 420}{2\times 1.40}=450nm$