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A surface of a prism having refractive index 1.5 is covered with a liquid of refractive index (3sqrt(2))/(4) . When be the minimum angle of incidence of an incident ray so that on the other surface of the prism the the ray will be totally reflected from the surface covered with liquid? The refracting angle of the prism75^(@) [sin48^(@)36' = 0.75]. |
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Answer» Solution :Let the critical angle between the prism and the LIQUID be `theta_(c)`. If the ray of light is totally REFLECTED from the surface covered with liquid then the angle of incidence of the ray on the surface is `theta_(c)` [Fig 2.63]. `1.5 sintheta_(c) = (3sqrt(2))/(4) sin90^(@) = (3sqrt(2))/(4)` `sintheta_c = (3sqrt(2))/(4) xx (10)/(15) = (1)/(sqrt(2))` `theta_(c) = 45^(@)` `"Again", r_(1) + theta_(c)= A or, r_(1) + 45^(@) = 75^(@) or, r_(1) = 30^(@)` `"Now" " " sini_(1) = musinr_(1) = 1.5 sin30^(@) = 0.75 or, i_(1) = 48^(@)36.` 
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