Question #134968
In a Young’s double slit interference experiment, the slits are at a distance 2L from each
other and the screen is at a distance D from the slits. If a glass slab of refractive index 
and thickness d is placed in the path of one of the beams, the minimum value of d for
the central fringe to be dark is
1
Expert's answer
2020-10-13T18:06:15-0400

Youngs double slit experiment

Δx=S\Delta x =S2PS*P-S 1*P

from pathegorian relationshipΔx=dsinθ\Delta x=d \sin \theta

sinθ\sin \theta =XnS\dfrac{Xn}{S}

\therefore Δx=\Delta x = XndS\dfrac{Xnd}{S}

from constructive interference Δx=mλ\Delta x =m \lambda

mλ=\therefore m \lambda = XndS\dfrac{Xn d}{S}

xn =mλd\dfrac{m \lambda}{d}


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