A Young's doublc-slit experiment is performed using monochromatic light of wavelength $\lambda$. The intensity of light at a point on the screen, where the path difference is $\lambda$, is $\mathrm{K}$ units. The intensity of light at a point where the path difference is $\mathrm{A} \frac{\lambda}{6}$ is given by $\frac{\mathrm{nK}}{12}$, where $\mathrm{n}$ is an integer. The value of $\mathrm{n}$ is
$\mathrm{I}_{\max }=\mathrm{k}$
$\mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{K} / 4$
$\Delta \mathrm{x}=\lambda / 6 \quad \Rightarrow \Delta \phi=\pi / 3$
$\mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}} \cos \phi$
$I=\frac{K}{4}+\frac{K}{4}+2 \times \frac{K}{4} \frac{1}{2}$
$=\frac{\mathrm{K}}{2}+\frac{\mathrm{K}}{4}=\frac{3 \mathrm{~K}}{4}=\frac{9 \mathrm{~K}}{12}$
$\mathrm{n}=9$
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