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 $420 nm$ and $630 nm$. 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$
$\Rightarrow t=\frac{2 \times 420}{2 \times 1.40} =450 \,nm$