InterviewSolution
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InterviewSolution
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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 Deltad=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. Find the thickness of the plate & wavelength of the incident light. |
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Answer» `rArr 0.25xx10^(-3)=(lambda)/(d)xx1 m` `rArr (lambda)/(d)=2.5xx10^(-4)…………..(1)` Afterwards : `(2)/(3) beta_(i) =(lambdaD)/((d+Deltad))` `rArr (lambda)/(d+Deltad)=(2)/(3)xx(2.5xx10^(-4))/(1)..............(2)` Dividing `(1) and (2)` `(d+Deltad)/(d)=(3)/(2) rArr d=2 (Deltad) =2.4 mm`. `& lambda=2.4xx2.5xx10^(-7)m = 600 nm`. Now `P` becomes central maxima. `rArr` for point `P : d sintheta =(mu-1)t` `rArr d. TANTHETA approx (mu-1)t` `rArr d(20beta)/(D) =(mu-1)t` `rArr (d)/(D)xx(20xxlambdaD)/(d) =(1.5-1)t` `rArr t=(20lambda)/(0.5) =(20xx600xx10^(-9))/(0.5)=24 MUM`. 
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