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} = \sqrt{A^{2} + A^{2} +2AA \cos \frac{\pi}{2}} = \sqrt{2 A^2} $
$\Rightarrow = I' = 2 I_0 \, \, \, \, \, \, \, \, \, $ ...(ii)
Required ratio = $ \frac{I'}{I} = \frac{2I_0}{4 I_0} = 0.5 $