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  • In the standard Young's double slit experiment, the intensity on the screen at a point distant 1.25 fringe widths from the central maximum is (assuming slits to be identical):<div class='qn

In the standard Young's double slit experiment, the intensity on the screen at a point distant 1.25 fringe widths from the central maximum is (assuming slits to be identical):

Option: 1

\mathrm{\frac{1}{2} I_{\max }}


Option: 2

\mathrm{\frac{1}{4} I_{\max }}


Option: 3

\mathrm{\frac{1}{3} I_{\max }}


Option: 4

\mathrm{I_{\max }}


Answers (1)

best_answer

Path difference, \mathrm{\Delta x=\frac{y d}{D}=\left(\frac{1.25 \lambda D}{d}\right)\left(\frac{d}{D}\right)=1.25 \lambda=\frac{5}{4} \lambda}

Path difference, \mathrm{\phi=\frac{2 \pi}{\lambda} \cdot \Delta x=\frac{2 \pi}{\lambda} \times \frac{5}{4} \lambda=\frac{5}{2} \pi}

\mathrm{ \begin{aligned} & \therefore \quad \frac{\phi}{2}=\frac{5}{4} \pi=225^{\circ} \\ & \therefore \quad \text { Intensity, } I=I_{\max } \cos ^2 \frac{\phi}{2}=\frac{1}{2} I_{\max } \end{aligned} }

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jitender.kumar

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