In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50

Question

In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness $t$ and refractive index $n(=1.5)$, the central fringe shifts by 0.2 cm . The value of $t$ is $\_\_\_\_$ cm.

Professor PC Thomas Classes · Answer

The shift in the central fringe due to the introduction of a transparent sheet is given by $\Delta x = \frac{D}{d}(n-1)t$. Substituting the given values: $0.2 	ext{ cm} = \frac{50 	ext{ cm}}{0.1 	ext{ cm}}(1.5 - 1)t \Rightarrow 0.2 = 500 	imes 0.5 	imes t \Rightarrow t = \frac{0.2}{250} 	ext{ cm} = 8 	imes 10^{-4} 	ext{ cm}$.