Question

Slit 1 of Young's double-slit experiment is wider than slit 2, so that the light from slits are given as $A_1=A_0\sin \omega t$ and $A_2=A_0\sin (\omega t+\frac{\pi }{3})$. The resultant amplitude and intensity, at a point where the path difference between them is zero, are $A$ and $I$ respectively. Then

• A. $A=\sqrt{3}{A}_0$
• B. $A=4A_0$
• C. $I\propto 16A_0^2$
• D. $I\propto 3A_0^2$

Answer 1

Answer: (A), (D)

The correct options are
A $A=\sqrt{3}{A}_0$
D $I\propto 3A_0^2$

$A_1=A_{0}sin\omega t$

and $A_2=A_{0}sin(\omega t+\frac{\pi }{3})=A_0\left[sin\omega tcos\right(\frac{\pi }{3})+cos\omega tsin(\frac{\pi }{3}\left)\right]=\frac{A_0}{2}(\sin \omega t+\sqrt{3}\cos \omega t)$

Resultant displacement=$A_1+A_2=A_0\frac{3}{2}\sin \omega t+A_0\frac{1}{2}\cos \omega t$

Let $Acos\varphi =\frac{3A_0}{2}$

$A\sin \varphi =\frac{A_0}{2}$

Therefore resultant displacement=$A\sin (\omega t+\varphi )$

$A=\sqrt{3}{A}_0$

$⟹I\propto 3A_0^2$