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
1 year, 8 months ago

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Sort by:

- 1 year, 8 months ago

Question 1- (B)?

- 1 year, 8 months ago

Question 2-(D)?

- 1 year, 8 months ago

@Sahil Silare can u write the solution

- 1 year, 8 months 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.

- 1 year, 8 months ago