Question

In Young's double-slit experiment, the intensity at a point is \(\left(\frac{1}{4}\right)\) of the maximum intensity. The angular position of this point is
\(\{\lambda\) = wavelength of light used, d = separation between the two slits, \(\cos (60)^{\circ }=\frac{1}{2}\}\)

• A. \(\sin ^{-1}\left(\frac{\lambda }{d}\right)\)
• B. \(\sin ^{-1}\left(\frac{\lambda }{2d}\right)\)
• C. \(\sin ^{-1}\left(\frac{\lambda }{3d}\right)\)
• D. \(\sin ^{-1}\left(\frac{\lambda }{4d}\right)\)

Hint:

Relate intensity to the phase difference, find the phase difference for one fourth intensity, and convert it into a path difference.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In Young's double slit experiment, \(I=I_{max}\cos^2\left(\dfrac{\phi}{2}\right)\), where \(\phi\) is the phase difference between the two waves.

Step 2: Find the phase difference
Given \(I=\dfrac{I_{max}}{4}\), so \(\cos^2\left(\dfrac{\phi}{2}\right)=\dfrac{1}{4}\) and \(\cos\dfrac{\phi}{2}=\dfrac{1}{2}\). Hence \(\dfrac{\phi}{2}=60^{\circ}\), which gives \(\phi=120^{\circ}=\dfrac{2\pi}{3}\).

Step 3: Calculation
Path difference \(=\dfrac{\lambda}{2\pi}\times\phi=\dfrac{\lambda}{2\pi}\times\dfrac{2\pi}{3}=\dfrac{\lambda}{3}\).
Since path difference \(=d\sin\theta\), we get \(\sin\theta=\dfrac{\lambda}{3d}\).

Final Answer:
The point lies at the angle whose sine is lambda over 3d, which is option (C). \[ \boxed{\text{(C) }\sin^{-1}\left(\dfrac{\lambda}{3d}\right)} \]