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  • Question Asked in: JEE Main-2015 In a Young’s double slit experiment with light of wavelength\lambda the separation of slits is d and distan

In a Young’s double slit experiment with light of wavelength\lambda the separation of slits is d and distance of screen is D such that D > > d > >\lambda. If the Fringe width is \beta, the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is :

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\begin{aligned} &\text {In YDSE } \Delta \mathrm{x} \text { is given as: } \Delta \mathrm{x}=\frac{\mathrm{yd}}{\mathrm{D}}\\ &\text {The fringe width is given as: } \beta=\frac{\lambda \mathrm{D}}{\mathrm{d}} \end{aligned}

\large \begin{array}{l} \text { Phase deff. } \phi=\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{\lambda} \cdot \frac{y d}{D} \Rightarrow \phi=\frac{2 \pi y}{\beta} \\ \text { Now } \frac{I_{\max }}{2}=I_{\max } \cos ^{2}\left(\frac{2 \pi y}{2 \beta}\right) \\ \Rightarrow \cos ^{2}\left(\frac{\pi y}{\beta}\right)=\frac{1}{2} \Rightarrow \frac{\pi y}{\beta}=\frac{\pi}{4} \Rightarrow y=\frac{\beta}{4} \end{array}

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