Question

In Young's double slit experiment, the light of wavelength '$\lambda$' is used. The intensity at a point on the screen is 'I' where the path difference is $\lambda/4$. If '$I_0$' denotes the maximum intensity then the ratio of '$I_0$' to 'I' is ($\cos 45^\circ = 1/\sqrt{2}$)

• A. 2 : 1
• B. 4 : 1
• C. 8 : 1
• D. 12 : 1

Hint:

At a path difference of $\lambda/4$, the phase difference is $90^\circ$, leading to exactly half the maximum intensity.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This problem relates path difference to phase difference and resultant intensity in an interference pattern.
Step 2: Key Formula or Approach:
1. Phase difference $\phi = \frac{2\pi}{\lambda} \Delta x$.
2. Resultant intensity $I = I_0 \cos^2\left(\frac{\phi}{2}\right)$.
Step 3: Detailed Explanation:
Given path difference $\Delta x = \frac{\lambda}{4}$.
Calculate phase difference $\phi$:
\[ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} = 90^\circ \]
Now calculate intensity $I$:
\[ I = I_0 \cos^2\left(\frac{\pi/2}{2}\right) = I_0 \cos^2\left(\frac{\pi}{4}\right) = I_0 \cos^2(45^\circ) \]
Using $\cos 45^\circ = \frac{1}{\sqrt{2}}$:
\[ I = I_0 \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2} \]
The ratio $\frac{I_0}{I}$ is:
\[ \frac{I_0}{I} = \frac{I_0}{I_0 / 2} = \frac{2}{1} \]
Step 4: Final Answer:
The ratio of $I_0$ to $I$ is 2 : 1.