In youngs double slit experiment the two equally bright slits are coherant but of phase differance pi/3.If the maximum intensity on screen is I.The intensity at the point on the screen equidistant from the slits is a)I b)I/2 c)I/4 d)3I/4 given ans is I/4 how ?

In interference of two light waves of intensity $I_1$ and $I_2$, the resultant intensity at a point is given by $I_R\:=\:I_1+I_2+2\sqrt{I_1 I_2}\:cos\:\phi$. Here $phi$ is the phase difference between the two light waves.

In YDSE, the two slits are identical. So the resultant intensity is given by $I_R\:=\:4I\:cos^2\phi$.

The maximum intensity produced is $4I$.

This maximum is given as $I_o$.

When two waves meet on the screen, the phase difference between them depends on two factors. First is the phase difference produced due to the difference in the path taken by the two waves.  Second is the initial phase difference when the wave starts at the slit.

When the waves meet at the midpoint on the screen path difference is zero. So the only phase difference is their initial phase difference.

SO the phase difference when the two waves meet at the centre of the screen sis $\frac{\pi}{3}$.

So, intensity at that point, $I_R\:=\:4I\:cos^2\phi\:=\:I_o\times cos^2\:(\frac{\pi}{3})\:=\:I_o\times \frac{1}{4}\:=\:\frac{I_o}{4}$.

So the required intensity is one-fourth of the maximum intensity on the screen