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(A) The deviation produced by a small-angled prism is given by $\delta_{1} = (\mu - 1) \alpha = (1.5 - 1) 	imes 4^o = 2^o$ (always).The deviation cau

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$1^o$ clockwise

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(A) The deviation produced by a small-angled prism is given by $\delta_{1} = (\mu - 1) \alpha = (1.5 - 1) 	imes 4^o = 2^o$ (always).The deviation caused by the mirror is $\delta_{2} = 180^o - 2i = 180^o - 2 	imes 45^o = 90^o$.The net deviation produced by the system is $\delta_{net} = \delta_{1} + \delta_{2} = 2^o + 90^o = 92^o$.Since the total deviation is $92^o$,which is greater than the required $90^o$,we need to decrease the deviation. The deviation produced by a mirror decreases as the angle of incidence increases.If the mirror is rotated by an angle $\beta$ in the clockwise direction,the new angle of incidence becomes $(45^o + \beta)$.The new deviation by the mirror is $\delta_{2}' = 180^o - 2(45^o + \beta) = 90^o - 2\beta$.The new total deviation is $\delta_{net}' = \delta_{1} + \delta_{2}' = 2^o + 90^o - 2\beta = 92^o - 2\beta$.Given that the total deviation should be $90^o$,we have $92^o - 2\beta = 90^o$,which gives $2\beta = 2^o$,or $\beta = 1^o$.Thus,the mirror should be rotated by $1^o$ in the clockwise direction.

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