Question

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is 'I'. The intensity at a point where the path difference is $\lambda/6$ is $[\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}][\lambda = \text{wavelength of light}][\cos \pi = -1]$

• A. I
• B. $\frac{3I}{4}$
• C. $\frac{1}{2}$
• D. $\frac{I}{4}$

Hint:

The general formula for intensity is $I = I_{max} \cos^2(\phi/2)$, where $\phi$ is the phase difference.

Answer 1

The Correct Option is B

Step 1: Reference Intensity
Path difference $\lambda \Rightarrow$ Phase difference $\phi = 2\pi$. Intensity $I = I_{max} \cos^2(2\pi/2) = I_{max} \cdot (1) = I_0$.
Step 2: Calculate New Phase Difference
Path difference $\Delta x = \lambda/6 \Rightarrow \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{\pi}{3}$.
Step 3: New Intensity Formula
$I' = I_{max} \cos^2(\phi/2) = I \cos^2(\frac{\pi/3}{2}) = I \cos^2(\frac{\pi}{6})$
Step 4: Calculation
$I' = I \cdot (\frac{\sqrt{3}}{2})^2 = I \cdot \frac{3}{4} = \frac{3I}{4}$
Final Answer:(B)