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The central fringe of the interference pattern produced by the light of wavelength 6000 Å is found to shift to the position of 4th dark fringe after a glass sheet of refractive index 1.5 is introduced. The thickness of glass sheet would beA. `4.8 mu m`B. `4.2 mu m`C. `5.4 mu m `D. `3.0 mu m ` |
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Answer» Correct Answer - B Shift `=((mu-1)tD)/d = 3.5, omega= (3.5lambdaD)/d` `t=(3.5lambda)/(mu-1)` `=((3.5)(6000 xx 10^-10))/(1.5 -1)` =`4.2 xx 10^(-6)m` =4.2 `mu`m. |
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