The refractive index of the material of the p | vedclass

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(B) The dispersive power $\omega$ of a prism is defined as the ratio of the angular dispersion to the mean deviation.It is given by the formula: $\ome

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$0.06$

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(B) The dispersive power $\omega$ of a prism is defined as the ratio of the angular dispersion to the mean deviation.It is given by the formula: $\omega = \frac{\mu_v - \mu_R}{\mu_y - 1}$Given:Refractive index for violet colour,$\mu_v = 1.69$Refractive index for red colour,$\mu_R = 1.65$Refractive index for mean colour (yellow),$\mu_y = 1.66$Substituting these values into the formula:$\omega = \frac{1.69 - 1.65}{1.66 - 1}$$\omega = \frac{0.04}{0.66}$$\omega = \frac{4}{66} = \frac{2}{33} \approx 0.0606$Rounding to two decimal places,we get $\omega = 0.06$.

The refractive index of the material of the prism for violet colour is $1.69$ and that for red is $1.65$. If the refractive index for mean colour is $1.66$,what is the dispersive power of the material of the prism?

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