InterviewSolution
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A monochromatic light of `lambda=500 nm` is incident on two identical slits separated by a distance of `5xx10^(-4)m`. The interference pattern is seen on a screen placed at a distance of `1 m` from the plane of slits. A thin glass plate of thickness `1.5xx10^(-6)m` and refractive index `mu=1.5` is placed between one of the slits and the screen. Find the intensity at the centre of the screen if the intensity is `I_(0)` in the absence of the plate. Also find the lateral shift of the central maxima and number of fringes crossed through centre. |
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Answer» `lambda=500 nm, d=5xx10^(-4) m, D=1 m` `t=1.5xx10^(-6)m, mu=1.5` Path difference due to plate, `Deltax=(mu-1)t=(1.5-1)xx1.5xx10^(-6)=0.75xx10^(-6)m` `phi=(2pi)/lambda Deltax=(2pi)/(500xx10^(-9))xx0.75xx10^(-6)=3pi` `I_(R)=I_(0) cos^(2) (phi//2)=I_(0) cos^(2)(1.5 pi)=0` Lateral shift `y_(0)=((mu-1)tD)/d=(0.75xx10^(-6)xx1)/(5xx10^(-4))` `=0.15xx10^(-2)m=1.5 mm` number of fringes crossing the centre `y_(0)/beta=(((mu-1)tD)/d)/((Dlambda)/d)=((mu-1)t)/lambda=(0.75xx10^(-6))/(500xx10^(-9))=1.5` |
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