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: Use amplitudes
Let each wave have amplitude $a$, so the maximum intensity is $I_0 = (2a)^2 = 4a^2$ (in suitable units).

Step 2: Resultant amplitude
For phase difference $\phi = 120^\circ$, the resultant amplitude is $2a\cos60^\circ = a$.

Step 3: Intensity
$I = a^2 = \dfrac{4a^2}{4} = \dfrac{I_0}{4}$.

Step 4: Answer
Option (A).

Step 5: Check with other path differences
For path difference 0 the intensity is $I_0$, for $\frac{\lambda}{4}$ it is $I_0\cos^2 45^\circ = \frac{I_0}{2}$, and for $\frac{\lambda}{2}$ it is zero. The value for $\frac{\lambda}{3}$ must lie between $\frac{I_0}{2}$ and zero, and $\frac{I_0}{4}$ does. The options $\frac{I_0}{2}$ and $I_0$ are the values for smaller path differences.

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