The glass prisms P_{1} and P_{2} are to be co | vedclass

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(D) For dispersion without deviation, the net deviation produced by the combination of two prisms must be zero.Let the refractive indices of prisms $P

Vedclass · Accepted Answer

$1.72$

Vedclass · Answer

(D) For dispersion without deviation, the net deviation produced by the combination of two prisms must be zero.Let the refractive indices of prisms $P_{1}$ and $P_{2}$ be $\mu$ and $\mu^{\prime}$ and their refracting angles be $A$ and $A^{\prime}$ respectively.The condition for no deviation is given by $\delta + \delta^{\prime} = 0$, which implies $(\mu - 1)A = (\mu^{\prime} - 1)A^{\prime}$.Given values are $\mu = 1.54$, $A = 4^{\circ}$, and $A^{\prime} = 3^{\circ}$.Substituting these values into the equation: $(1.54 - 1) 	imes 4^{\circ} = (\mu^{\prime} - 1) 	imes 3^{\circ}$.$0.54 	imes 4 = (\mu^{\prime} - 1) 	imes 3$.$2.16 = (\mu^{\prime} - 1) 	imes 3$.$\mu^{\prime} - 1 = \frac{2.16}{3} = 0.72$.$\mu^{\prime} = 0.72 + 1 = 1.72$.

The glass prisms $P_{1}$ and $P_{2}$ are to be combined together to produce dispersion without deviation. The angles of the prisms $P_{1}$ and $P_{2}$ are selected as $4^{\circ}$ and $3^{\circ}$ respectively. If the refractive index of prism $P_{1}$ is $1.54,$ then that of $P_{2}$ will be

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