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Question

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A. $\dfrac{\omega }{\delta }$

B. $\delta /s$

C. $\dfrac{1}{{\omega \delta }}$

D. $\omega \delta $

Answer

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Here, in the question it is given that the dispersive power of the prism is $\omega $.

And the angle of deviation is $\delta $ .

Angular dispersion = angular deviation of violet – angular deviation of red

i.e. Angular dispersion = ${\delta _V} - {\delta _R}$

Dispersive power is given by the ratio of angular dispersion to the angle of deviation of the mean colour between the two extremes (I.e. yellow).

\[

{\text{Dispersive power = }}\dfrac{{{\text{Angular dispersion}}}}{{{\text{mean deviation}}}} \\

\Rightarrow \omega = \dfrac{{{\delta _V} - {\delta _R}}}{{{\delta _{mean}}}} - - - - - - - - - - - - - [1] \\

\]

From eq.1 we can find that:

Angular dispersion = ${\delta _V} - {\delta _R} = \omega {\delta _{mean}}$

The angle of deviation given in the question must be mean deviation (i.e. ${\delta _{mean}} = \delta $ )

Hence, the answer is: Angular dispersion = $\omega \delta $

Angular dispersion is the difference in deviations of the two extreme frequencies in the spectrum produced by the prism. Dispersive power of the prism is given by the ratio of angular dispersion to the mean angle of deviation. So, we can find the formula of angular dispersion from the above formula. Using this information, we can find out the correct answer.