Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

The condition for interference of light waves depends on the path difference ($\Delta x$) between the two waves:

• For Constructive Interference (Bright Fringe): $\Delta x = n\lambda$, where $n = 0, 1, 2, 3, \dots$
• For Destructive Interference (Dark Fringe): $\Delta x = (2n - 1)\frac{\lambda}{2}$, where $n = 1, 2, 3, \dots$

Given data:

• Path difference at $P_1$: $\Delta x_1 = \frac{3\lambda}{2}$
• Path difference at $P_2$: $\Delta x_2 = 2\lambda$

Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]

Evaluating the conditions for each point:

• For point $P_1$: $\Delta x_1 = \frac{3\lambda}{2}$. This matches the condition for destructive interference, $(2n - 1)\frac{\lambda}{2}$ (where $n=2$). Thus, a dark fringe is formed at $P_1$.
• For point $P_2$: $\Delta x_2 = 2\lambda$. This matches the condition for constructive interference, $n\lambda$ (where $n=2$). Thus, a bright fringe is formed at $P_2$.

Step 3 (Final Concluded Answer): [0.25 Mark]

Comparing the results with the given options, we conclude that a bright fringe is formed at $P_2$ and a dark fringe is formed at $P_1$.

Correct Option: (iv)

💡 EL10 Concept Insight:

Think of light waves like two people walking; if they arrive at the same time (whole number of wavelengths), they boost each other to make a bright spot, but if one is "half a step" behind (odd multiple of half-wavelengths), they cancel each other out to create darkness.