Question

In an interference experiment using waves of same amplitude, path difference between the waves at a point on the screen is $\frac{\lambda }{4}$. The ratio of intensity at this point with that at the central bright fringe is

• A. 1
• B. 0.5
• C. 1.5
• D. 2.0

Answer 1

Answer: (B)

Let amplitude of each wave = $A$
and its corresponding intensity = $I_0$
Phase difference between waves at central bright fringe = 0.
So, resultant amplitude,·
$A_R=\sqrt{A^2+A^2+2AA\cos 0}$
$=\sqrt{4A^2}$
$A_R^2=4A^2$
or , $I=4I_0$
Path difference between the waves at a point on the screen = $\frac{\lambda }{4}$
Corresponding phase difference $=\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}$
So, resultant amplitude at that point
$A_R^{\prime }=\sqrt{A^2+A^2+2AA\cos \frac{\pi }{2}}=\sqrt{2A^2}$
$\Rightarrow =I^{\prime }=2I_0$ ...(ii)
Required ratio = $\frac{I^{\prime }}{I}=\frac{2I_0}{4I_0}=0.5$