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A ray of light incident normally on a refracting surface of the prism is totally reflected from the other refracting surface. If the prism is immersed in water how will the ray act? Refractive index of glass = 1.5, refractive index of water = 1.33. |
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Answer» Solution :As the ray of light is incident on a FACE of the prism normally, so it goes straight through the surface. The ray is incident on the second refracting face and is TOTALLY reflected. So the angle of incidence of the ray at the second face is greater than the critical angle `(theta_(c))`. `sintheta_(c) = (1)/(a^(mu)g) = (1)/(1.5) = 0.667` `=sin41.8^(@)` `THEREFORE "" theta_(c) = 41.8^(@)` Now, if the prism is immersed in water, refractive index of glass with respect to water is, `w^(mu)g = (a^(mu)g)/(a^(mu)w) = (1.5)/(1.33) = 1.128` In this case if the critical angle is `theta_(c)`, then `sintheta._(c) = (1)/(w^(mu)g) = (1)/(1.128) = 0.8865` `= sin 62.44^(@)` `or, "" theta._(c) = 62.44^(@)` So the angle of incidence of the ray at the second face `(41.8^(@))` is lessthan the critical angle `(62.44^(@))`. Hence, instead of being totally reflected from the second face, the ray is REFRACTED throughit. |
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