Question #182699

Derive an expression for the displacement of the nth bright fringe in Young’s

double-slit experiment when a thin transparent plate of refractive index  and

thickness t is introduced in the path of one of the constituent interfering beams of

light. Will there be any change in the fringe-width after the introduction of the

plate?


1
Expert's answer
2021-05-05T16:30:35-0400

When a thin transparent plate of thickness ‘t’ is introduced in front of one of the slits in Young’s double slit experiment, the fringe pattern shifts toward the side where the plate is present.





The dotted lines denote the path of the light before introducing the transparent plate. The solid lines denote the path of the light after introducing a transparent plate.


Path difference before introducing the plate Δx=S1PS2P\Delta x=S_1P-S_2P


Path Difference after introducing the plate Δxnew=S1P1S2P1\Delta x_{new}=S_1P_1-S_2P_1


The path length S2P1=(S2P1t)air+tplate=(S2P1tair)+(μt)plateS_2P_1=(S_2P_1-t)_{air}+t_{plate}=(S_2P_1-t_{air})+(\mu t)_{plate}

where 'μt\mu t ' is optical path.

=S2P1air+(μ1)t=S_2P_{1air}+(\mu -1)t


Then we get


(Δx)new=S1P1air(S2P1air+(μ1)t)=(S1P1S2P1)air(μ1)t(Δx)new=dsinθ(μ1)t(Δx)new=γdD(μ1)t(\Delta x)_{new}=S_1P_{1air}-(S_2P_{1air}+(\mu -1)t)=(S_1P_1-S_2P_1)_{air}-(\mu-1)t\\(\Delta x)_{new}=dsin\theta-(\mu-1)t(\Delta x)_{new}=\frac{\gamma d}{D}-(\mu-1)t\\

Then,

y=ΔxDd+Dd[(μ1)t]y=\frac{\Delta x D}{d}+\frac{D}{d}[(\mu-1)t]


The term ΔxDd\frac{\Delta x D}{d} defines the position of a bright or dark fringe, the term Dd[(μ1)t]\frac{D}{d}[(\mu-1)t] defines the shift occurred in the particular fringe due to the introduction of a transparent plate.


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