Question

In biprism experiment the maximum intensity is ' \(I_0\) '. If the path difference between the two interfering waves is ' \(\lambda/4\) ' then intensity at the point on the screen is \(\left[ \sin 45^\circ = \cos 45^\circ = \frac{1}{\sqrt{2}} \right]\)

• A. \(\frac{I_0}{4}\)
• B. \(\frac{I_0}{3}\)
• C. \(\frac{I_0}{2}\)
• D. \(I_0\)

Hint:

Formula: $I = I_0 \cos^2(\delta/2)$

Answer 1

The Correct Option is C

Concept: Intensity in interference: \[ I = I_0 \cos^2\left(\frac{\delta}{2}\right) \]

Step 1: Phase difference. \[ \delta = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2} \]

Step 2: Substitute. \[ I = I_0 \cos^2\left(\frac{\pi}{4}\right) \] \[ \cos 45^\circ = \frac{1}{\sqrt{2}} \Rightarrow I = I_0 \times \frac{1}{2} \]

Step 3: Conclusion.
$I = \frac{I_0}{2}$ Final Answer: Option (C)