Question

In Young's double slit experiment, the intensity at a point where the path difference is $\frac{\lambda}{4}$ ($\lambda$ is wavelength of light used) is '$I$'. If '$I_0$' is the maximum intensity then $I/I_0$ is equal to

• A. $3:2$
• B. $2:3$
• C. $3:4$
• D. $1:2$

Hint:

Remember the standard fractional path differences: $\lambda/2$ gives minimum intensity ($0$), $\lambda$ gives maximum ($I_0$), and $\lambda/4$ gives exactly half the maximum intensity ($I_0/2$).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the path difference at a specific point on the screen in a Young's double slit experiment. We need to find the ratio of the intensity at that point ($I$) to the maximum central intensity ($I_0$).

Step 2: Key Formula or Approach:
First, convert the path difference ($\Delta x$) to phase difference ($\phi$) using the relation $\phi = \frac{2\pi}{\lambda} \Delta x$.
Next, use the intensity formula for interference: $I = I_0 \cos^2\left(\frac{\phi}{2}\right)$.

Step 3: Detailed Explanation:
Given path difference $\Delta x = \frac{\lambda}{4}$.
Calculate the phase difference $\phi$:
$$\phi = \frac{2\pi}{\lambda} \times \left(\frac{\lambda}{4}\right) = \frac{\pi}{2}$$
Now, substitute this phase difference into the intensity formula:
$$I = I_0 \cos^2\left(\frac{\pi / 2}{2}\right) = I_0 \cos^2\left(\frac{\pi}{4}\right)$$
Since $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$, squaring it gives $\frac{1}{2}$:
$$I = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = I_0 \left(\frac{1}{2}\right) = \frac{I_0}{2}$$
Therefore, the ratio is:
$$\frac{I}{I_0} = \frac{1}{2}$$

Step 4: Final Answer:
The ratio $I/I_0$ is $1:2$, matching option (D).