A thin prism with angle 5^{circ} and refracti | vedclass

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(C) For a thin prism, the deviation produced is given by $\delta = (\mu - 1)A$.For dispersion without deviation, the net deviation must be zero, i.e.,

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(C) For a thin prism, the deviation produced is given by $\delta = (\mu - 1)A$.For dispersion without deviation, the net deviation must be zero, i.e.,$\delta_{net} = \delta_{1} + \delta_{2} = 0$.Since the prisms are combined to produce dispersion without deviation, they must be placed in opposite orientations, so $(\mu_{1} - 1)A_{1} + (\mu_{2} - 1)A_{2} = 0$.Taking the magnitude, we have $(\mu_{1} - 1)A_{1} = -(\mu_{2} - 1)A_{2}$.Substituting the given values: $(1.72 - 1) 	imes 5^{\circ} = -(1.9 - 1) 	imes A_{2}$.$0.72 	imes 5^{\circ} = -0.9 	imes A_{2}$.The negative sign indicates the orientation of the second prism. The magnitude of the angle $A_{2}$ is:$A_{2} = \frac{0.72 	imes 5^{\circ}}{0.9} = \frac{3.6^{\circ}}{0.9} = 4^{\circ}$.

$A$ thin prism with angle $5^{\circ}$ and refractive index $1.72$ is combined with another prism of refractive index $1.9$ to produce dispersion without deviation. The angle of the second prism is . . . . . . . (in $^{\circ}$)

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