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Minimum and maximum deviation in a prism A 60^(@) prism has a refractive index of 1.5. (a) Calculate the angle of incidence for minimum deviation. |
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Answer» Solution :As discussed in the Section : Angle of Minimum Deviation, the angle of incidence will be EQUAL to the angle of emergence . Calculation : As the prism is in the position of minimum deviation , `r-A//2=60^(@)//2-30^(@)`, so that at either face `sini=1.5sin30^(@)=0.75` `i=sin^(-1)(0.75)=49^(@)` (B) Calculate the angle of emergence of light at maximum deviation As we have descussed in the Section : Angle of maximum Deviation, the angle of incidence will be equal to `90^(@)`. Calculation :For maximum deviation `i_(1)=90^(@)` so that `r_(1)=theta_(C )=sin^(-1)(2//3)=42^(@)`. But as in a prism. `r_(1)+r_(2)=A` Therefore, `r_(2)=A-r_(1)=60^(@)-42^(@)=18^(@)` Now applying Snell.s law at the second face, `nsinr_(2)=sini_(2)`, `(3)/(2)sin18^(@)=sini_(2)` that is `i_(2)=sin^(-1)(1.5xx0.31)=sin^(-1)(0.465)=28^(@)` |
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