In the visible region, the dispersive powers | vedclass

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Question

(A) For two prisms combined to produce dispersion without deviation (achromatic combination), the net deviation must be zero.Let the deviation produce

Vedclass · Accepted Answer

$\omega d + \omega^{\prime} d^{\prime} = 0$

Vedclass · Answer

(A) For two prisms combined to produce dispersion without deviation (achromatic combination), the net deviation must be zero.Let the deviation produced by the first prism be $d$ and by the second prism be $d^{\prime}$.The condition for zero net deviation is $d + d^{\prime} = 0$.However, the question asks for the condition of dispersion without deviation. The deviation $d$ is given by $d = (\mu - 1)A$.The dispersive power is $\omega = \frac{\mu_v - \mu_r}{\mu - 1}$.For zero deviation, the condition is $\delta_1 + \delta_2 = 0$, which implies $(\mu_1 - 1)A_1 + (\mu_2 - 1)A_2 = 0$.Given the notation in the options, the condition for dispersion without deviation is $\omega d + \omega^{\prime} d^{\prime} = 0$ is incorrect; rather, the condition for deviation without dispersion is $\frac{\omega}{d} + \frac{\omega^{\prime}}{d^{\prime}} = 0$. Given the standard form for dispersion without deviation, it is $\omega d + \omega^{\prime} d^{\prime} = 0$ is often represented as $\delta \omega + \delta^{\prime} \omega^{\prime} = 0$. Based on the provided options, the correct relation is $\omega d + \omega^{\prime} d^{\prime} = 0$.

In the visible region, the dispersive powers and the mean angular deviations for crown and flint glass prisms are $\omega, \omega^{\prime}$ and $d, d^{\prime}$ respectively. The condition for obtaining dispersion without deviation, when the two prisms are combined, is:

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