MHTCET 13 Apr 2026 Shift 2 Paper

Concept:In Young’s double slit experiment, the $n$th bright fringe has a path difference of $n\lambda$.Explanation:Take point $O$ as the central maximum, where path difference is zero.$Q$ is the second bright fringe on the right, so its path difference is:$$\Delta x_Q = 2\lambda$$$P$ is the eleventh bright fringe on the other side measured from $Q$.So, moving 11 fringe positions from $Q$ toward the left gives:$$\Delta x_P = 2\lambda - 11\lambda = -9\lambda$$The negative sign shows that $P$ is on the opposite side of $O$.Hence the magnitude of the path difference at $P$ is $9\lambda$.From the figure, $S_2P \approx BP$ because the screen is far away.Therefore, the path difference at $P$ is represented by $S_1B$:$$S_1B = 9\lambda$$Given:$$\lambda = 6000\ \mathring{A} = 6000 \times 10^{-10}\ \mathrm{m} = 6 \times 10^{-7}\ \mathrm{m}$$Substitute the value:$$S_1B = 9 \times 6 \times 10^{-7} = 5.4 \times 10^{-6}\ \mathrm{m}$$Answer:D. $5.4 \times 10^{-6}\ \mathrm{m}$