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Refractive index of a prism A ray is incident on an equilateral prism such that the angle of deviation is 30^(@) . It is seen that if the angle of incidence is increased by 30^(@), then the deviation is again 30^(@). Find the refractive index of the prism. |
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Answer» Solution :We can EASILY see that in one SITUATION the angle of incidence will be `i` and in the next situation angle of incidence will be `i+30^(@)`. From the above DISCUSSION, we can SAY that in the first situation, the angle of emergence should be `i+30^(@)` and in the second situation, the angle of emergence should be `i`. Calculation : In both the cases `delta=i+e-A` `30^(@)=i+i+30^(@)-60^(@)` `i=30^(@)` The angle of emergence thus is `i+30^(@)=60^(@)` By Snell.s law, `1sin30=nsinr_(1)` By Snell.s law at the second surface we have `1sin60=nsinr_(2)` `1sin60=nsin(60-r_(1))` Solving for n we GET `n=sqrt((4+sqrt(3))/(3))`. |
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