A thin film of soap solution $(\mathrm{n}=1.4)$ lies on the top of a glass plate $(\mathrm{n}=1.5)$. When visible light is incident almost normal to the plate, two adjacent reflection maxima are observed at two wavelengths 420 and $630 \mathrm{~nm}$. The minimum thickness of the soap solution is

$\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ are the two rays considered for interference. $R_{1}$ is the result of reflection at denser medium; hence it suffers an additional path difference of $\frac{\lambda}{2}$. Ray $R_{2}$ originates after reflection at narer medium. Net path difference $=2 \mu \mathrm{t}+\frac{\lambda}{2}$ where $t$ is the thickness of the soap solution. For constructive interference, $2 \mu \mathrm{t}+\frac{\lambda}{2}=\mathrm{m} \lambda$, where $\mathrm{m}=0,1,2 \ldots . .$ Let the two adjacent reflection maxima be observed at $\mathrm{m}$ and $\mathrm{m}-1$. Then $2 \mu \mathrm{t}+\frac{\lambda_{1}}{2}=\mathrm{m} \lambda_{1}$ or $2 \mu \mathrm{t}=\left(\mathrm{m}-\frac{1}{2}\right) \lambda_{1}$ ...(1) Similarly, $2 \mu \mathrm{t}=\left(\mathrm{m}-\frac{3}{2}\right) \lambda_{2}$ ...(2) Solving (1) \& (2), we get
$2 \mu \mathrm{t}=\left(\frac{\lambda_{2}}{\frac{\lambda_{2}}{\lambda_{1}}-1}\right)$ Putting, $\lambda_{1}=420 \mathrm{~nm}, \lambda_{2}=630 \mathrm{~nm}, \mu=1.4$ We get, $t=450 \mathrm{~nm}$