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: Understanding the Question:
In YDSE, the resultant intensity at a point depends on the phase difference between the two waves, which is directly proportional to the path difference.
Step 2: Key Formula or Approach:
1. Phase difference $\phi = \frac{2\pi}{\lambda} \Delta x$.
2. Intensity $I_{res} = I_{max} \cos^2\left(\frac{\phi}{2}\right)$.
Step 3: Detailed Explanation:
Case 1: Path difference $\Delta x = \lambda$.
\[ \phi_1 = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi \]
Resultant intensity:
\[ I = I_{max} \cos^2\left(\frac{2\pi}{2}\right) = I_{max} \cos^2(\pi) = I_{max} \cdot (-1)^2 = I_{max} \]
So, the maximum intensity is $I$.
Case 2: Path difference $\Delta x' = \frac{\lambda}{6}$.
\[ \phi_2 = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{\pi}{3} = 60^\circ \]
Resultant intensity $I'$:
\[ I' = I_{max} \cos^2\left(\frac{\phi_2}{2}\right) = I \cos^2\left(\frac{\pi/3}{2}\right) = I \cos^2\left(\frac{\pi}{6}\right) \]
Given $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$:
\[ I' = I \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = I \cdot \frac{3}{4} = \frac{3I}{4} \]
Step 4: Final Answer:
The intensity is $\frac{3I}{4}$.