In a YDSE experiment, the distance between the slits & the screen is 100 cm. For a certain distance between the slits, an interference pattern is observed on the screen with the fringe width 0.25 mm. When the distance between the slits is increased by Δd = 1.2 mm, the fringe width decreased to n = 2/3 of the original value. In the final position, a thin glass plate of refractive index 1.5 is kept in front of one of the slits & the shift of central maximum is observed to be 20 fringe width. The thickness of the plate in the form of 12x μm. Fill 'x'
-
UP 0 DOWN 0 0 1
1 Answers
1\text{Fringe Width}=\frac{\lambda D}{d} \\ \lambda=(0.25*10^{-3})d \\ d_{new}=d+1.2*10^{-3} \\ \text{New Fringe Width}=\frac{2\lambda}{3d}=\frac{\lambda}{d+(1.25*10^{-3})}\\ \\ \text{Now Solve For d and }\lambda ...\\ \text{I Hope U Can Solve The Question N}
