Question

In YDSE, intensity at path difference $\lambda/3$ is I. Find intensity at path difference $\lambda$.

• A. 2I
• B. 3I
• C. 4I
• D. I

Hint:

Max intensity occurs at path difference = integer multiples of $\lambda$.

Answer 1

The Correct Option is D

Step 1: Intensity in YDSE: \[ I = I_0(1 + \cos\phi) \]
Step 2: Phase difference: \[ \phi = \frac{2\pi}{\lambda}\Delta x \]
Step 3: For $\Delta x = \lambda/3$: \[ \phi = \frac{2\pi}{3} \] \[ I = I_0(1 + \cos120^\circ) = I_0(1 - 1/2) = I_0/2 \]
Step 4: For $\Delta x = \lambda$: \[ \phi = 2\pi \Rightarrow \cos\phi = 1 \Rightarrow I = 2I_0 \]
Step 5: Ratio: \[ I_{\lambda} : I_{\lambda/3} = 2I_0 : I_0/2 = 4:1 \] Hence intensity at $\lambda$ is 4I.