Physics - Wave Optics Question with Solution | TestHub

If an interference experiment is performed using two wavelengths close to each other, two interference patterns corresponding to the two wavelengths are obtained on the screen. The fringe system remains distinct upto a point on the screen where the$n^{th}$order maximum of  one wavelength, say${\lambda }_1=5890 \dot{\mathrm{A}}$falls on the$n^{th}$order minimum of the order wavelength${\lambda }_2=5895\dot{ \mathrm{A}}$. Thus, interference pattern can be seen upto a distance$y_n$from the centre of the screen if

$y_n=\frac{n{\lambda }_{1}D}{d}; \left(n^{\mathrm{th}}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{x}\mathrm{i}\mathrm{m}\mathrm{u}\mathrm{m}\right)\dots \dots \dots ..\left(i\right)$

$=\left(n-\frac{1}{2}\right)\frac{{\lambda }_{2}D}{d};\left(n^{\mathrm{th}}\mathrm{ }\mathrm{m}\mathrm{a}\mathrm{x}\mathrm{i}\mathrm{m}\mathrm{u}\mathrm{m}\right)\dots \dots \dots \left(ii\right)$

Or$n{\lambda }_1=\left(n-\frac{1}{2}\right){\lambda }_2 or 2n{\lambda }_1=\left(2n-1\right){\lambda }_2$

This gives,

$n=\frac{{\lambda }_2}{2\left({\lambda }_2-{\lambda }_1\right)}$, which is choice$n=\frac{{\lambda }_2}{2\left({{\lambda }_2-\lambda }_1\right)}$