Q. At two points $P$ and $Q$ on screen in Young's double slit experiment. Waves from slits $S_{1}$ and $S_{2}$ have a path difference$\ of\ 0\ and\frac{\lambda}{4}respectively.\ \ $The ratio of intensities at P and Q will be
A) $3\ :2$
B) $2\ :1$
C) $\sqrt{2}\ :1$
D) $4\ :1$
Explanation
(b) Let $I_{0}$ is intensity of light emitted from the source, then
Resultant intensity
$I = 4I_{0}\cos^{2}\frac{\phi}{2}\ $
$I_{1} = 4I_{0}\cos^{2}\frac{\phi}{2} = 4I_{0}$
Now, $\ \ \ \ \ \ \mathrm{\Delta}x = \frac{\lambda}{4}$
$\phi = \frac{2\pi}{\lambda} \times \mathrm{\Delta}x = \frac{2\pi}{\lambda} \times \frac{\lambda}{4}$
$\phi = \frac{\pi}{2}$
And $\ I_{2} = 4I_{0}\cos^{2}{\frac{\pi}{4} = 2I_{0}}$
$I_{1}:I_{2} = 2\ :1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.