Question

In YDSE, the intensity where path difference is (\frac{\lambda}{4}) is (\frac{K}{4}). The intensity at a point when path difference is ' (\lambda) ' will be

• A. (4 K)
• B. (2 K)
• C. (K)
• D. [suspicious link removed]

Hint:

Maximum intensity occurs when the path difference is an integral multiple of $\lambda$.

Answer 1

The Correct Option is D

Step 1: Intensity Formula
$I = I_{max} \cos^2\left(\frac{\Delta \phi}{2}\right)$, where $\Delta \phi = \frac{2\pi}{\lambda} \Delta x$.

Step 2: First Condition
For $\Delta x = \lambda/4$, $\Delta \phi = \pi/2$. $I_1 = I_{max} \cos^2(\pi/4) = \frac{I_{max}}{2}$. Given $I_1 = K/4$, so $I_{max} = K/2$.

Step 3: Second Condition
For $\Delta x = \lambda$, $\Delta \phi = 2\pi$. $I_2 = I_{max} \cos^2(\pi) = I_{max} = \frac{K}{2}$.
Final Answer: (D)