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Figure shows a YDSE setup having identical slits `S_(1)` and `S_(2)` with d =5 mm and D = 1 m. A monochromatic light of wavelength `lamda = 6000 Å` is incident on the plane of slit due to which at screen centre O, an intensity `I_(0)` is produced with fringe pattern on both sides Now a thin transparent film of `11 mu m` thickness and refractive index `mu = 2.1` is placed in front of slit `S_(1)` and now interference patten is observed again on screen. After placing the film of slit `S_(1)`, the intensity at point O screen is :A. 0.1 mmB. 0.06 mmC. 0.02 mmD. None of th |
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Answer» Correct Answer - C After introducing the film, path difference at point O is `Delta = t(mu-1)=11 xx10^(-6)xx1.1 = 12.1 xx10^(-6)m` Phase difference at point O is `phi=(2 pi)/(lamda)xx Delta` `=(2pi)/(6 xx 10^(-7))12.1 xx 10^(-6)` `=(12.1pi)/(3)=(pi)/(3)` Thus intensity at point O is `I=I_(0) "cos"^(2)(phi)/(2)=I_(0)"cos"^(2)(pi)/(6)= 3 (I_(0))/(4)` Now path difference at point in terms of `lamda` is `Delta _(0)=(12.1 xx10^(-6))/(6xx 10^(-7))lamda = 20.167 lamda` Thus due to placement of thin film 20 bright fringes cross the point O. |
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