Physics - Modern Physics Question with Solution | TestHub

${\mathrm{R}}_1$ 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 $\mathrm{m}=0,1,2\dots ..$ 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, $t=450\mathrm{nm}$