Wave Optics Part 2

Concept:The shift of the central fringe in YDSE occurs due to an additional optical path difference introduced by a transparent sheet.Explanation:Optical path through the sheet is $nt$.Path in air for same distance is $t$.Extra optical path difference: $\Delta x_{\text{optical}} = (n-1)t$.Geometric path difference for a point at distance $y$ from centre: $\Delta x_{\text{geo}} = \frac{d y}{D}$.At the shifted central maximum, net path difference is zero:$\frac{d y}{D} = (n-1)t$.Thus, $t = \frac{y d}{D (n-1)}$.Given: $d = 0.1$ cm, $D = 50$ cm, $y = 0.2$ cm, $n = 1.5$.Substitute: $t = \frac{0.2 \times 0.1}{50 \times (1.5 - 1)} = \frac{0.02}{50 \times 0.5} = \frac{0.02}{25}$.Calculate: $t = 8 \times 10^{-4}$ cm.Answer:$8 \times 10^{-4}$ cm, which corresponds to option B.