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  • In a Young's double slit experiment, D equals the distance of screen and d is the separation between the slits. The distance of the nearest point to the centre maximum where the intensity is sa

In a Young's double slit experiment, D equals the distance of screen and d is the separation between the slits. The distance of the nearest point to the centre maximum where the intensity is same as that due to a single slit, is equal to:

Option: 1

\mathrm{\frac{\mathrm{D} \lambda}{\mathrm{d}}}


Option: 2

\mathrm{\frac{\mathrm{D} \lambda}{2 \mathrm{~d}}}


Option: 3

\mathrm{\frac{\mathrm{D} \lambda}{3 \mathrm{~d}}}


Option: 4

\mathrm{\frac{2 \mathrm{D} \lambda}{\mathrm{d}}}


Answers (1)

best_answer

Intensity, \mathrm{I=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}}

\mathrm{ \begin{aligned} & \because \quad \mathrm{I}=\mathrm{I}_0 \\ & \therefore \quad \phi=\frac{2 \pi}{3}=\frac{2 \pi}{\lambda} \cdot \Delta \mathrm{x}=\frac{2 \pi}{\lambda}\left(\frac{\mathrm{yd}}{\mathrm{D}}\right) \\ & \therefore \quad \mathrm{y}=\frac{\lambda \mathrm{D}}{3 \mathrm{~d}} \end{aligned} }

Posted by

Kuldeep Maurya

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