Question

In Young's double slit experiment, the intensity on screen at a point, where path difference is \(\frac{\lambda}{4}\) is \(\frac{K}{4}\). The intensity at a point when path difference is ' \(\lambda\) ' will be \(\left[\cos \frac{\pi}{2}=0, \cos 2 \pi=1\right]\)

• A. 4 K
• B. 2 K
• C. K
• D. \(\frac{\mathrm{K}}{2}\)

Hint:

Use I = I0 cos^2(phase/2) with phase = 2 pi times path difference over lambda, find I0 from the first case.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Intensity at a point is \(I=I_{0}\cos^{2}\dfrac{\phi}{2}\), where \(\phi=\dfrac{2\pi}{\lambda}\Delta\) and \(I_{0}\) is the maximum intensity.

Step 2: First case
\(\Delta=\dfrac{\lambda}{4}\Rightarrow\phi=\dfrac{\pi}{2}\), so \(I=I_{0}\cos^{2}\dfrac{\pi}{4}=\dfrac{I_{0}}{2}=\dfrac{K}{4}\).
Hence \(I_{0}=\dfrac{K}{2}\).

Step 3: Second case
\(\Delta=\lambda\Rightarrow\phi=2\pi\), so \(I=I_{0}\cos^{2}\pi=I_{0}=\dfrac{K}{2}\).

Final Answer:
The intensity at path difference lambda is K/2. \[ \boxed{\text{(D) }\dfrac{\mathrm{K}}{2}} \]