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In a Young’s double slit experiment, using unequal slit widths, the intensity at a point midway between a bright and dark fringes is 4 I. If one slit is covered by an opaque film, intensity at that point becomes 2 I. If the other is covered instead, then the intensity at that point is:

Option: 1

\mathrm{2 \mathrm{I}}


Option: 2

\mathrm{5 \mathrm{I}}


Option: 3

\mathrm{(5+2 \sqrt{2}) \mathrm{I}}


Option: 4

\mathrm{(5+4 \sqrt{2}) \mathrm{I}}


Answers (1)

best_answer

\mathrm{\text { Location of the point, } y=\frac{\omega}{4}=\frac{\lambda D}{4 d}}

\mathrm{\begin{aligned} & \Rightarrow \Delta \mathrm{x}=\frac{\mathrm{yd}}{\mathrm{D}}=\frac{\lambda}{4} \\\\ & \Rightarrow \quad \phi=\frac{2 \pi}{\lambda} \cdot \Delta \mathrm{x}=\frac{\pi}{2} \\\\ & \because \mathrm{I}_{\mathrm{R}}=\mathrm{I}_1+\mathrm{I}_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \phi \\\\ & \therefore \quad 4 \mathrm{I}=2 \mathrm{I}+\mathrm{I}_2+0 \\\\ & \Rightarrow \mathrm{I}_2=2 \mathrm{I} \end{aligned}}

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SANGALDEEP SINGH

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