Question

At two points \(P\) and \(Q\) on the screen in Young's double slit experiment, waves from slits \(S_{1}\) and \(S_{2}\) have a path difference of zero and \(\frac{\lambda }{4}\) respectively. The ratio of intensities at \(P\) and \(Q\) will be \(\left(\cos 0^{\circ }=1,\cos 45^{\circ }=\frac{1}{\sqrt{2}}\right)\)

• A. \(3:2\)
• B. \(2:1\)
• C. \(\sqrt{2}:1\)
• D. \(4:1\)

Hint:

Convert each path difference to a phase difference and use I = 4I0 cos^2 of half the phase difference.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
In Young's experiment with equal-intensity slits, the intensity is \(I=4I_0\cos^2\left(\frac{\phi}{2}\right)\), where the phase difference is \(\phi=\frac{2\pi}{\lambda}\Delta x\).

Step 2: Point P
Path difference is 0, so \(\phi=0\).
\(I_P=4I_0\cos^2 0^{\circ}=4I_0\).

Step 3: Point Q
Path difference is \(\frac{\lambda}{4}\), so \(\phi=\frac{\pi}{2}\).
\(I_Q=4I_0\cos^2 45^{\circ}=4I_0\times\frac{1}{2}=2I_0\).

Step 4: Ratio
\(\frac{I_P}{I_Q}=\frac{4I_0}{2I_0}=2:1\), which is option (B).

Final Answer:
The ratio of intensities at P and Q is 2:1. \[ \boxed{\text{(B) }2:1} \]