Question

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}

• A. 4 K
• B. 2 K
• C. K
• D. $\frac{K}{4}$

Hint:

Path difference of $\lambda$ corresponds to constructive interference (maximum intensity).

Answer 1

The Correct Option is C

Step 1: Concept
Intensity $I$ at a point is given by $I = I_{max} \cos^2(\phi/2)$, where phase difference $\phi = \frac{2\pi}{\lambda} \times \text{path difference} (\Delta x)$.

Step 2: Meaning
For $\Delta x = \lambda/4$, $\phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2}$. Intensity $I_1 = I_{max} \cos^2(\pi/4) = I_{max}(1/2) = K/2$. Thus, $I_{max} = K$.

Step 3: Analysis
For $\Delta x = \lambda$, $\phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi$. Intensity $I_2 = I_{max} \cos^2(2\pi/2) = I_{max} \cos^2(\pi)$.

Step 4: Conclusion
Since $\cos^2(\pi) = 1$, $I_2 = I_{max} = K$. Final Answer: (C)