Question

In a Young's double slit experiment, the intensity at a point where the path difference is \( \frac{\lambda}{6} \) (\( \lambda \) being the wavelength of light used) is \( I \). If \( I_{0} \) denotes the maximum intensity, \( I/I_{0} \) is equal to:

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

Hint:

Always remember: Path difference of \(\lambda\) corresponds to a phase difference of \(2\pi\). This makes it easy to visualize: \(\lambda/2 \to \pi\) (minima), \(\lambda/4 \to \pi/2\), etc.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
In wave interference, the intensity at any point depends on the phase difference between the two waves. The path difference provided by the setup can be converted into a phase difference.

Step 2: Key Formula or Approach
1. Phase difference (\(\phi\)) = \(\frac{2\pi}{\lambda} \times \Delta x\) (where \(\Delta x\) is path difference).
2. Intensity \(I = I_0 \cos^2(\phi/2)\).

Step 3: Detailed Explanation
1. Find Phase Difference: Given \(\Delta x = \lambda/6\). \[ \phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{2\pi}{6} = \frac{\pi}{3} = 60^\circ \]
2. Calculate Intensity Ratio: \[ I = I_0 \cos^2\left(\frac{60^\circ}{2}\right) = I_0 \cos^2(30^\circ) \] \[ \cos 30^\circ = \frac{\sqrt{3}}{2} \implies \cos^2(30^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \]
3. Result: \[ \frac{I}{I_0} = \frac{3}{4} \]

Step 4: Final Answer
The ratio \(I/I_0\) is 3/4.