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In an experiment for determination of refractive index of glass of a prism by `i-delta`, plot it was found thata ray incident at angle `35^circ`, suffers a deviation of `40^circ` and that it emerges at angle `79^circ`. In that case which of the following is closest to the maximum possible value of the refractive index?A. `1.5`B. `1.6`C. `1.7`D. `1.8` |
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Answer» Correct Answer - A Here `i=35^(@)`, `e=79^(@)`, `delta=40^(@)` we know `delta=i+e-AimpliesA=i+e-delta` `A=35^(@)+79^(@)-40^(@)=74^(@)` Refractive index of prism `mu=(sin((A+delta_(m))/(2)))/(sin((A)/(2)))` `mu=(sin(37^(@)+(delta_(m))/(2)))/(sin37^(@))=(5)/(3)sin(37^(@)+(delta_(m))/(2))` Maximum value of `mu` can be `(5)/(3)`, so required value of `mu` should be less than `(5)/(3)` also `delta_(m)` will be less than `40^(@)`, so `mult(5)/(3)sin(37^(@)+(40^(@))/(2))=(5)/(2)sin57^(@)` `mult(5)/(3)sin57^(@)lt(5)/(3)sin60^(@)=1.44` `mult1.44` so the nearest possible value of `mu` for the given arrangement should be `1.5` |
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