A thin prism of angle 6^{circ} made of glass | vedclass

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Question

(C) For dispersion without deviation, the net deviation produced by the combination must be zero.$\delta - \delta^{\prime} = 0 \Rightarrow A(\mu - 1)

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) For dispersion without deviation, the net deviation produced by the combination must be zero.$\delta - \delta^{\prime} = 0 \Rightarrow A(\mu - 1) - A^{\prime}(\mu^{\prime} - 1) = 0$Where:$A = 6^{\circ}$ (angle of the first prism)$\mu = 1.5$ (refractive index of the first prism)$A^{\prime} = ?$ (angle of the second prism)$\mu^{\prime} = 1.75$ (refractive index of the second prism)Substituting the values into the equation:$A^{\prime}(\mu^{\prime} - 1) = A(\mu - 1)$$A^{\prime}(1.75 - 1) = 6(1.5 - 1)$$A^{\prime}(0.75) = 6(0.5)$$A^{\prime} = \frac{6 	imes 0.5}{0.75} = \frac{3}{0.75} = 4^{\circ}$Thus, the angle of the second prism is $4^{\circ}$.

$A$ thin prism of angle $6^{\circ}$ made of glass of refractive index $1.5$ is combined with another prism made of glass of refractive index $1.75$ to produce dispersion without deviation. Then, the angle of the second prism is (in $^{\circ}$)

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