Question:medium

In Young's double slit experiment, the intensity on screen at a point where path difference is $\frac{\lambda}{4}$ is $\frac{K}{2}$. The intensity at a point when path difference is '$\lambda$' will be}

Show Hint

Path difference of $\lambda$ corresponds to constructive interference (maximum intensity).
Updated On: May 14, 2026
  • 4 K
  • 2 K
  • K
  • $\frac{K}{4}$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understanding the Concept:
The intensity in an interference pattern varies with the phase difference between the two waves.
The phase difference is derived from the path difference.
We determine the maximum possible intensity ($I_{\text{max}}$) from the first condition and use it to find the intensity for the second condition.
Step 2: Key Formula or Approach:
Relation between Phase difference $\phi$ and Path difference $\Delta x$: $\phi = \frac{2\pi}{\lambda} \Delta x$.
Intensity formula: $I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right)$.
Step 3: Detailed Explanation:
Case 1: Path difference $\Delta x_1 = \frac{\lambda}{4}$.
Calculate phase difference $\phi_1$: \[ \phi_1 = \frac{2\pi}{\lambda} \left(\frac{\lambda}{4}\right) = \frac{\pi}{2} \text{ radians} \] The intensity $I_1 = K/2$. Using the intensity formula: \[ I_1 = I_{\text{max}} \cos^2\left(\frac{\pi/2}{2}\right) = I_{\text{max}} \cos^2\left(\frac{\pi}{4}\right) \] Since $\cos(\pi/4) = 1/\sqrt{2}$, its square is $1/2$. \[ \frac{K}{2} = I_{\text{max}} \left(\frac{1}{2}\right) \implies I_{\text{max}} = K \] Case 2: New path difference $\Delta x_2 = \lambda$.
Calculate new phase difference $\phi_2$: \[ \phi_2 = \frac{2\pi}{\lambda} (\lambda) = 2\pi \text{ radians} \] Calculate new intensity $I_2$: \[ I_2 = I_{\text{max}} \cos^2\left(\frac{2\pi}{2}\right) = I_{\text{max}} \cos^2(\pi) \] Since $\cos(\pi) = -1$, its square is $1$. \[ I_2 = I_{\text{max}} \cdot 1 = I_{\text{max}} \] We found $I_{\text{max}} = K$, therefore $I_2 = K$.
Step 4: Final Answer:
The intensity is K.
Was this answer helpful?
0