Question

In YDSE, phase difference between the interfering waves at a point on screen having intensity less than the average intensity on screen may be :

• A. $\frac{\pi }{4}$
• B. $\frac{2\pi }{3}$
• C. $\pi$
• D. $\frac{7\pi }{8}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $\frac{2\pi }{3}$
C $\pi$
D $\frac{7\pi }{8}$

We know that intensity at any point is given by -
$I_p=I_{0}co{s}^2\left(\frac{\varphi }{2}\right)$
$I_0$ = maximum intesity
$\varphi$= phase difference
Average intensity $I_{avg}=\frac{I_0}{2}$
It is given in the question that
$I_P<\frac{I_0}{2}$
So, $I_0{\text{cos}}^2\frac{\varphi }{2}<\frac{I_0}{2}.$
${\text{cos}}^2\frac{\varphi }{2}<\frac{1}{2}\Rightarrow \text{cos}\frac{\varphi }{2}\text{ranges from}(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$
$\frac{\pi }{4}<\frac{\varphi }{2}<\frac{3\pi }{4}$
$\frac{\pi }{2}<\varphi <\frac{3\pi }{2}$