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In an isosceles prism of angle `45^@`, it is found that when the angle of incidence is same as the prism angle, the emergen ray grazes the emergent surface. Find the refractive index of the material of the prism. For what angle of incidenc, the angle of deviation will be minimum? |
Answer» As the ray of light grazes the second surface, `r_(2)` is the critical angle. `sinr_(2)=(1)/(mu)` `r_(2)=(45^@-r_(1))` `sinr_(1)=(sin45^@)/(mu)=(1)/(sqrt(2)mu)` `sinr_(2)=sin(45^@-r_(1))` `=(1)/(sqrt(2))[cosr_(1)-sinr_(1)]` `(1)/(mu)=(1)/(sqrt(2))[sqrt(1-(1)/(2mu^(2)))-(1)/(sqrt(2)mu)]` `(1)/(mu)=(1)/(sqrt(2))[sqrt(2mu^(2)-1)/(sqrt(2)mu)-(1)/(sqrt(2)mu)]` `2=sqrt(2mu^(2)-1)-1` `2mu^(2)-1=9` `2mu^(2)=10` `rArr mu^(2)=5` or `mu=sqrt(5)` At minimum deviation, `r_(1)=r_(2)(45^(@))/(2)=22.5^(@)` `mu=(sini_(1))/(sinr_(1))rArrsini_(1)=(sqrt(5))sin(22.5^(@))` `i_(1)=58.8^(@)` |
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