A Young's double-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 K units. The intensity of light at a point where the path difference is $\frac{\lambda }{6}$ is given by $\frac{nK}{12}$ , where n is an integer. The value of n is __________
Option: 1 9
Option: 2 6
Option: 3 3
Option: 4 12

Answers (1)

D Deependra Verma

$\begin{array}{l} \mathrm{I}_{\max }=\mathrm{k} \\\\ \mathrm{I}_{1}=\mathrm{I}_{2}=\mathrm{K} / 4 \\\\ \Delta \mathrm{x}=\lambda / 6 \Rightarrow \Delta \phi=\pi / 3 \\\\ \mathrm{I}=\mathrm{I}_{1}+\mathrm{I}_{2}+2 \sqrt{\mathrm{I}_{1} \mathrm{I}_{2}} \cos \phi \\\\ \mathrm{I}=\frac{\mathrm{K}}{4}+\frac{\mathrm{K}}{4}+2 \times \frac{\mathrm{K}}{4} \frac{1}{2} \\\\ =\frac{\mathrm{K}}{2}+\frac{\mathrm{K}}{4}=\frac{3 \mathrm{~K}}{4}=\frac{9 \mathrm{~K}}{12} \\\\ \mathrm{n}=9 \end{array}$

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