WBJEE 2026 Physics and Chemistry Paper

Concept:Two coherent sources produce circular interference fringes on the screen.The path difference is maximum at $O$ and decreases as we move away from $O$, because $D \gg d$.Explanation:At point $O$, the path difference is $\Delta x = d$.For maxima: $\Delta x = m\lambda$.For minima: $\Delta x = (2m+1)\frac{\lambda}{2}$.If $d = \frac{7\lambda}{2}$, then $\Delta x = \frac{7\lambda}{2} = (2 \times 3 + 1)\frac{\lambda}{2}$.So $O$ is a minima.Hence option A is correct.If $d = \lambda$, then at $O$, $\Delta x = \lambda = 1 \cdot \lambda$.So $O$ is a maxima, not a minima.As the path difference decreases from $\lambda$ to $0$, the only maximum on the finite screen is at $O$.Hence option B is correct and option D is incorrect.If $d = 4.8\lambda$, then minima occur for$$\Delta x = \frac{\lambda}{2},\ \frac{3\lambda}{2},\ \frac{5\lambda}{2},\ \frac{7\lambda}{2},\ \frac{9\lambda}{2}$$All these values are less than $4.8\lambda$.So there are 5 minima on the screen.Hence option C is correct.Answer:Correct statements are A, B and C.