The figure shows a YDSE set-up. The source S of wavelength \(\lambda = 4000 A^{\circ}\) oscillates along the y-axis according to the equation \(y = \sin(\pi t)\) where \(y\) is in millimetres and \(t\) is in seconds. The distance between the two slits is \(0.5 \text{ mm}\). Answer the following questions:

**Q.** The instant at which maximum intensity occurs at \(P\) for the first time is:

\(\dfrac{1}{\pi} \arcsin \left( \dfrac{59}{160} \right)\)

\(\dfrac{1}{\pi} \arcsin \left( \dfrac{59}{80} \right)\)

\(\dfrac{1}{\pi} \arcsin \left( \dfrac{27}{160} \right)\)

\(\dfrac{1}{\pi} \arcsin \left( \dfrac{27}{80} \right)\)

**Q.** The instant at which minimum intensity occurs at \(P\) for the first time is:

\(\dfrac{1}{\pi} \arcsin \left( \dfrac{59}{160} \right)\)

\(\dfrac{1}{\pi} \arcsin \left( \dfrac{59}{80} \right)\)

\(\dfrac{1}{\pi} \arcsin \left( \dfrac{27}{160} \right)\)

\(\dfrac{1}{\pi} \arcsin \left( \dfrac{27}{80} \right)\)

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## Comments

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TopNewest@Steven Chase @Mark Hennings @Md Zuhair Please help.

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Question 1- (B)?

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Question 2-(D)?

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@Sahil Silare can u write the solution

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I'm bad with latex but what I did was adding up the path difference from source to slits and slits to screen and then it will be equal to n(lambda) for maximum.

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