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

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(A) For dispersion without deviation,the net deviation produced by the combination must be zero.The condition for deviation without dispersion is give

Vedclass · Accepted Answer

$4^{\circ} 30^{\prime}$

Vedclass · Answer

(A) For dispersion without deviation,the net deviation produced by the combination must be zero.The condition for deviation without dispersion is given by $\delta_1 + \delta_2 = 0$.For thin prisms,the deviation is $\delta = (\mu - 1)A$.Thus,$(\mu_1 - 1)A_1 + (\mu_2 - 1)A_2 = 0$.Given $A_1 = 6^{\circ}$,$\mu_1 = 1.54$,and $\mu_2 = 1.72$.Substituting the values: $(1.54 - 1) 	imes 6^{\circ} + (1.72 - 1) 	imes A_2 = 0$.$0.54 	imes 6^{\circ} + 0.72 	imes A_2 = 0$.$3.24^{\circ} + 0.72 	imes A_2 = 0$.$A_2 = -\frac{3.24}{0.72} = -4.5^{\circ}$.The negative sign indicates that the prism $P_2$ must be inverted relative to $P_1$. The magnitude of the angle is $4.5^{\circ}$,which is $4^{\circ} 30^{\prime}$.

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

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