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. Then 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

The correct option is C $\frac{\pi }{3}$

The correct answer is option(C).

We know, The resultant intensity,

$I_R=4I_{max}co{s}^2\left(\frac{\varphi }{2}\right)$

$or,co{s}^2\left(\frac{\varphi }{2}\right)=\frac{I_R}{4I_{m}ax}=\frac{0.75}{4}=0.1875$

$\Rightarrow cos\left(\frac{\varphi }{2}\right)=\sqrt{0.1875}=0.5$

$\Rightarrow \frac{\varphi }{2}=co{s}_{-1}\left(0.5\right)=\frac{\pi }{3}$