Question

In the Young's double slit experiment, the intensity of light at a point on the screen where the path difference is \(\lambda\) is 'K', \(\lambda\) being the wavelength of light used. The intensity at a point where path difference is \(\frac{\lambda }{6}\) will be \(\left[\cos \left(\frac{\pi }{6}\right)=\frac{\sqrt{3}}{2}\right]\)

• A. \(\frac{K}{6}\)
• B. \(\frac{K}{4}\)
• C. \(\frac{K}{2}\)
• D. \(\frac{3K}{4}\)

Hint:

A path difference of one wavelength is a bright fringe, so K is the maximum intensity.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
In the double slit experiment, the intensity is \(I=I_0\cos^{2}\left(\dfrac{\phi}{2}\right)\), where \(\phi\) is the phase difference and \(I_0\) is the maximum intensity.

Step 2: Find the Maximum Intensity
Path difference \(\lambda\) gives \(\phi=2\pi\), a bright fringe. So \(I_0=K\).

Step 3: Calculation
For path difference \(\dfrac{\lambda}{6}\), \(\phi=\dfrac{2\pi}{\lambda}\times\dfrac{\lambda}{6}=\dfrac{\pi}{3}\). Then \(I=K\cos^{2}\left(\dfrac{\pi}{6}\right)=K\times\dfrac{3}{4}\).

Final Answer:
The intensity at that point is \(\dfrac{3K}{4}\). \[ \boxed{\text{(D) }\dfrac{3K}{4}} \]