Q.Two light sources of equal amplitudes interfere with each other. Calculate the ratio of maximum and minimum intensities.

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 …