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: Understanding the Question:
The intensity in an interference pattern depends on the phase difference between the two waves, which in turn is related to the path difference.
Step 2: Key Formula or Approach:
1) Phase difference \(\delta = \frac{2\pi}{\lambda} \times \text{Path difference}(\Delta x)\).
2) Resultant intensity \(I = I_{max} \cos^2\left(\frac{\delta}{2}\right)\).
Step 3: Detailed Explanation:
Case 1: Path difference \(\Delta x_1 = \lambda/4\).
Phase difference \(\delta_1 = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}\).
Intensity \(I_1 = \frac{K}{4} = I_{max} \cos^2\left(\frac{\pi/2}{2}\right) = I_{max} \cos^2\left(\frac{\pi}{4}\right)\).
\[ \frac{K}{4} = I_{max} \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{I_{max}}{2} \]
\[ I_{max} = \frac{2K}{4} = \frac{K}{2} \]
Case 2: Path difference \(\Delta x_2 = \lambda\).
Phase difference \(\delta_2 = \frac{2\pi}{\lambda} \times \lambda = 2\pi\).
Intensity \(I_2 = I_{max} \cos^2\left(\frac{2\pi}{2}\right) = I_{max} \cos^2(\pi)\).
Since \(\cos \pi = -1\), \(\cos^2 \pi = 1\).
\[ I_2 = I_{max} \times 1 = I_{max} \]
Substituting the value of \(I_{max}\) from Case 1:
\[ I_2 = \frac{K}{2} \]
Step 4: Final Answer:
The intensity will be \(\frac{K}{2}\).