Question

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

• A. \(4K\)
• B. \(2K\)
• C. \(7K\)
• D. \(K/2\)

Hint:

Use the intensity formula for two equal coherent sources and find the maximum intensity from the first given point.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For two equal slits the intensity at a point is \(I = I_{max}\cos^{2}(\phi/2)\), where the phase difference is \(\phi = \dfrac{2\pi}{\lambda}\times\) path difference.

Step 2: Using the first point
Path difference \(\lambda/4\) gives \(\phi = \pi/2\).
So \(I = I_{max}\cos^{2}(\pi/4) = I_{max}/2\).
This is given as \(K/4\), so \(I_{max} = K/2\).

Step 3: Using the second point
Path difference \(\lambda\) gives \(\phi = 2\pi\).
So \(\cos^{2}(\pi) = 1\) and \(I = I_{max} = K/2\).

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