A thin prism P_1 with an angle 6^{circ} and m | vedclass

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(D) For dispersion without average deviation,the net deviation produced by the combination must be zero.The condition for no average deviation is give

Vedclass · Accepted Answer

$4.5$

Vedclass · Answer

(D) For dispersion without average deviation,the net deviation produced by the combination must be zero.The condition for no average deviation is given by $\delta_1 + \delta_2 = 0$,which implies $|\delta_1| = |\delta_2|$.The deviation produced by a thin prism is $\delta = A(\mu - 1)$.For prism $P_1$: $A_1 = 6^{\circ}$,$\mu_1 = 1.54$.For prism $P_2$: $A_2 = A$,$\mu_2 = 1.72$.Equating the deviations: $A_1(\mu_1 - 1) = A_2(\mu_2 - 1)$.$6^{\circ}(1.54 - 1) = A(1.72 - 1)$.$6^{\circ}(0.54) = A(0.72)$.$A = \frac{6^{\circ} 	imes 0.54}{0.72} = \frac{6 	imes 54}{72} = \frac{324}{72} = 4.5^{\circ}$.

$A$ thin prism $P_1$ with an angle $6^{\circ}$ and made of glass of refractive index $1.54$ is combined with another prism $P_2$ made from glass of refractive index $1.72$ to produce dispersion without average deviation. The angle of prism $P_2$ is $........{}^{\circ}$

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