A thin prism P_1 with angle 4^o made of glass | vedclass

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(A) For a thin prism,the deviation $\delta$ is given by $\delta = A(\mu - 1)$.Since the combination produces no net deviation,the deviation produced b

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$3$

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(A) For a thin prism,the deviation $\delta$ is given by $\delta = A(\mu - 1)$.Since the combination produces no net deviation,the deviation produced by prism $P_1$ must be equal and opposite to the deviation produced by prism $P_2$.Let $A_1 = 4^o$,$\mu_1 = 1.54$ and $A_2 = ?$,$\mu_2 = 1.72$.The condition for no deviation is $\delta_1 + \delta_2 = 0$,which implies $A_1(\mu_1 - 1) + A_2(\mu_2 - 1) = 0$.Substituting the values: $4(1.54 - 1) + A_2(1.72 - 1) = 0$.$4(0.54) + A_2(0.72) = 0$.$2.16 + A_2(0.72) = 0$.$A_2 = -2.16 / 0.72 = -3^o$.The negative sign indicates that the prism $P_2$ must be inverted relative to $P_1$. The magnitude of the angle is $3^o$.

$A$ thin prism $P_1$ with angle $4^o$ made of glass of refractive index $1.54$ is combined with another thin prism $P_2$ made of glass of refractive index $1.72$ to produce no deviation. The angle of prism $P_2$ is.....$^o$

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