Question

In the Young's double slit experiment, the resultant intensity at a point on the screen is $75\%$ of the maximum intensity of the bright fringe. The phase difference between the two interfering rays at that point is:

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{2}$

Answer 1

Answer: (C)

$\frac{\pi }{3}$

The intensity at a point with phase difference $\alpha$ between two interfering rays is given by the formula,

$I^{\prime }=I+I+2Icos\left(\alpha \right)$

$I_{max}=I+I+2Icos\left(0\right)=4I$

$I_P=\frac{3}{4}4I=2I+2Icos\alpha$

$1+cos\alpha =\frac{3}{2}$

$cos\alpha =0.5$

$\alpha =\frac{\pi }{3}$

Answer is option C.