Q.Two light waves from slit $S_1$ and $S_2$ on reaching points P and Q on a screen in Young's double slit experiment have a path difference zero and $\lambda/4$ respectively. The ratio of light intensities at P and Q will be :

Interference is the redistribution of light intensity that results when two light waves overlap (are superposed) at a point, producing bright regions (increased intensity) at some points and dark regions (decreased intensity) at others, in a way that a single wave alone never would. If two waves of the same frequency and amplitudes $a_1,a_2$ arrive at a point with a constant phase difference $\phi$ between them, $y_1=a_1\sin\omega t$ and $y_2=a_2\sin(\omega t+\phi)$, their resultant is $y=A\sin(\omega t+\theta)$ with resultant amplitude $A^2=a_1^2+a_2^2+2a_1a_2\cos\phi$; since intensity $I\propto A^2$, the resultant intensity is $I\propto I_1+I_2+2\sqrt{I_1I_2}\cos\phi$. Constructive interference (maximum brightness) occurs when $\phi=0,\pm2\pi,\pm4\pi,\ldots$, giving $I_{\max}\propto(a_1+a_2)^2\propto I_1+I_2+2\sqrt{I_1I_2}$; destructive interference (minimum brightness, possibly complete darkness) occurs when $\phi=\pm\pi,\pm3\pi,\ldots$, giving $I_{\min}\propto(a_1-a_2)^2\propto I_1+I_2-2\sqrt{I_1I_2}$. For the special case of equal amplitudes ($a_1=a_2=a$, $I_1=I_2=I_0$), this simplifies …