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A `60^@` prism has a refractive index of `1.5`. Calculate (a) the angle of incidence for minimum deviation (b) angle of minimum deviation ( c) the angle of emergence of light at maximum deviation (d) angle of maximum deviation. |
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Answer» Here, `A=60^(@), mu=1.5` When deviation is minimum , `r=(A)/(2)=(60)/(2)=30^(@)` As `mu=(sin i)/(sin i)` `sini=mu sin r=1.5 sin 30^(@)=0.75` `i=sin^(-1)(0.75)=49^(@)` (b) `del_(m)=2i-A=2xx49^(@)-60^(@)=38^(@)` (c) For maximum deviation, `i_(1)=90^(@)` and `r_(1)=C=sin^(=1)((1)/(mu))=sin^(-1)((1)/(3//2))=sin^(-1)(0.6667)=42^(@)` As `r_(1)+r_(2)=A` `r_(2)=A-r_(1)=60-42^(@)=18^(@)` From `mu=(sini_(2))/(sinr_(2)),` `sini_(2)=mu sin r_(2)=1.5 sin 18^(@)=1.5xx0.31=0.4650` `i_(2)=sin^(-1)(0.4650)=28^(@)` This is the angle of emergent (d) `del_("max")=i_(1)+i_(2)-A=90^(@)+28^(@)-60^(@)=58^(@)` |
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