Question

A thin film of soap solution (${\mu }_s=1.4$) lies on the top of a glass plate (${\mu }_g=1.5$). When visible light is incident 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. $420nm$
• B. $500nm$
• C. $450nm$
• D. $490nm$

Answer 1

The correct option is C

$450nm$

Step 1: Given Data:

Refractive index of soap solution, ${\mu }_s=1.4$

Refractive index of glass, ${\mu }_g=1.5$

Two adjacent reflection maxima are observed at two wavelengths,

$$\begin{gathered} {\lambda }_1=420nm \\ {\lambda }_2=630nm \end{gathered}$$

Let the thickness of the soap solution is $t$.

Step 2: We have to find the minimum thickness of the soap solution is,

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 $\pi$.

The condition for maximum intensity:

$2\mu t=n\lambda$ (where, $\mu$is the refractive index and $n$ is order)

Step 3: For order $n$:

As we know, the formula of refractive index is $n_1{\lambda }_1=n_2{\lambda }_2$

For n,

$n{\lambda }_1=(n-1){\lambda }_2$

$\Rightarrow$ $n\times 420=(n-1)630$

$\Rightarrow$ $n\times 420=n\times 630-630$

$\Rightarrow$$n\times 420-n\times 630=-630$

$\Rightarrow$ $n(420-630)=-630$

$\Rightarrow$ $n(-210)=-630$

$\Rightarrow$ $210n=630$

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

$\Rightarrow$ $n=3$

Step 4: Maximum order, where they coincide:

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

$\Rightarrow$ $t=\frac{2\times 1.4}{3\times 420}$

$\Rightarrow$ $t=450nm$

Hence, the correct option is (C).