Question

In biprism experiment, the maximum intensity is \(I_0\). If the path difference between the two interfering waves is \(\frac{λ}{3}\), then intensity at the point on the screen is
[\(sin30^{\circ} = cos60^{\circ} = 0.5\), \(sin60^{\circ} = cos30^{\circ} = \sqrt{3}/2\)]

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

Hint:

Convert path difference to phase difference and use I = I0 cos squared of half the phase.

Answer 1

The Correct Option is A

Step 1: Understand the concept
A path difference \(\Delta x\) corresponds to a phase difference \(\phi = \dfrac{2\pi}{\lambda}\Delta x\). For two equal sources, the intensity is \(I = I_0\cos^2\dfrac{\phi}{2}\), where \(I_0\) is the maximum intensity.

Step 2: Find the phase difference
\[ \phi = \frac{2\pi}{\lambda}\cdot\frac{\lambda}{3} = \frac{2\pi}{3} \]

Step 3: Compute
\[ I = I_0\cos^2\frac{\pi}{3} = I_0\cos^2 60^\circ = I_0\left(\frac{1}{2}\right)^2 = \frac{I_0}{4} \]

Step 4: Result
The intensity is \(\dfrac{I_0}{4}\), option (A). The value \(I_0/2\) would be at a path difference of \(\lambda/4\), and \(I_0\) occurs at zero path difference.

Final Answer:
The intensity is I0/4. This is option (A). \[ \boxed{\text{(A) }\frac{I_0}{4}} \]