A prism with a refractive index of 1.5 and an | vedclass

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(C) The deviation produced by the prism is given by $\delta_{prism} = (\mu - 1)A$.Substituting the given values: $\delta_{prism} = (1.5 - 1) 	imes 4^

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$178$

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(C) The deviation produced by the prism is given by $\delta_{prism} = (\mu - 1)A$.Substituting the given values: $\delta_{prism} = (1.5 - 1) 	imes 4^o = 0.5 	imes 4^o = 2^o$.The ray strikes the mirror at an angle of incidence $i = \delta_{prism} = 2^o$ relative to the normal (since the ray is incident normally on the first face of the prism).The deviation produced by the mirror is $\delta_{mirror} = 180^o - 2i = 180^o - 2(2^o) = 180^o - 4^o = 176^o$.The total deviation is the sum of the deviations produced by the prism and the mirror: $\delta_{total} = \delta_{prism} + \delta_{mirror} = 2^o + 176^o = 178^o$.

$A$ prism with a refractive index of $1.5$ and an angle $A = 4^o$ is placed in front of a mirror. When the incident ray is reflected by the mirror,what is the total deviation in degrees?

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