Compare the dispersive powers of two prisms i | vedclass

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(C) The dispersive power $\omega$ of a prism is defined as $\omega = \frac{\delta_v - \delta_r}{\delta_y}$,where $\delta_v$ and $\delta_r$ are the dev

Vedclass · Accepted Answer

$0.89$

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(C) The dispersive power $\omega$ of a prism is defined as $\omega = \frac{\delta_v - \delta_r}{\delta_y}$,where $\delta_v$ and $\delta_r$ are the deviations of violet (or blue) and red rays,and $\delta_y$ is the mean deviation,given by $\delta_y = \frac{\delta_v + \delta_r}{2}$.For the first prism: $\delta_v = 10^{\circ}$,$\delta_r = 6^{\circ}$.$\delta_{y1} = \frac{10^{\circ} + 6^{\circ}}{2} = 8^{\circ}$.$\omega_1 = \frac{10^{\circ} - 6^{\circ}}{8^{\circ}} = \frac{4}{8} = 0.5$.For the second prism: $\delta_v = 8^{\circ}$,$\delta_r = 4.5^{\circ}$.$\delta_{y2} = \frac{8^{\circ} + 4.5^{\circ}}{2} = 6.25^{\circ}$.$\omega_2 = \frac{8^{\circ} - 4.5^{\circ}}{6.25^{\circ}} = \frac{3.5}{6.25} = 0.56$.The ratio of dispersive powers is $\frac{\omega_1}{\omega_2} = \frac{0.5}{0.56} \approx 0.89$.

Compare the dispersive powers of two prisms if one of them deviates the blue and red rays through $10^{\circ}$ and $6^{\circ}$ respectively,and the second prism through $8^{\circ}$ and $4.5^{\circ}$.

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