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:

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

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

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

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

Note by Tapas Mazumdar
3 months, 1 week ago

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Question 2-(D)?

- 3 months, 1 week ago

@Sahil Silare can u write the solution

- 3 months, 1 week ago

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.

- 3 months, 1 week ago

Question 1- (B)?

- 3 months, 1 week ago