Question

In Young's double slit experiment using a monochromatic light of wavelength \(\lambda\). The path difference (in terms of an integer n) corresponding to any point having one fourth of its peak intensity is
\(\{\cos (\pi /3)=1/2,\ \cos (4\pi /3)=-1/2\}\)

• A. \(\left(n\pm \frac{1}{2}\right)\lambda ,\ n=0,1,2\)
• B. \(\left(n\pm \frac{1}{3}\right)\lambda ,\ n=0,1,2\)
• C. \(\left(n\pm \frac{1}{4}\right)\lambda ,\ n=0,1,2\)
• D. \(\left(n\pm \frac{1}{5}\right)\lambda ,\ n=0,1,2\)

Hint:

Write intensity as a cosine-squared function of the phase difference and set it equal to one fourth of the maximum.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For two equal slits the intensity is \(I=I_0\cos^{2}\dfrac{\phi}{2}\), where \(I_0\) is the peak intensity and \(\phi\) is the phase difference.

Step 2: Set up the condition
We need \(I=\dfrac{I_0}{4}\), so \(\cos^{2}\dfrac{\phi}{2}=\dfrac{1}{4}\) and \(\cos\dfrac{\phi}{2}=\pm\dfrac{1}{2}\).

Step 3: Find the phase difference
Using the given values \(\cos(\pi/3)=\tfrac12\) and \(\cos(4\pi/3)=-\tfrac12\), we get \(\dfrac{\phi}{2}=\dfrac{\pi}{3}\) or \(\dfrac{4\pi}{3}\) (plus multiples of \(\pi\)), so \(\phi=2n\pi\pm\dfrac{2\pi}{3}\).

Step 4: Convert to path difference
\(\Delta x=\dfrac{\lambda}{2\pi}\phi=\left(n\pm\dfrac{1}{3}\right)\lambda\).

Final Answer:
The path difference is \(\left(n\pm\tfrac13\right)\lambda\). \[ \boxed{\text{(B) }\left(n\pm \dfrac{1}{3}\right)\lambda ,\ n=0,1,2} \]