The refractive index of a particular material | vedclass

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

Vedclass · Accepted Answer

$0.0615$

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(A) The dispersive power $\omega$ of a material 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 values:Refractive index for blue light (violet end),$\mu_v = 1.67$Refractive index for red light,$\mu_r = 1.63$Refractive index for yellow light (mean),$\mu_y = 1.65$Substituting these values into the formula:$\omega = \frac{1.67 - 1.63}{1.65 - 1}$$\omega = \frac{0.04}{0.65}$$\omega = \frac{4}{65} \approx 0.0615$Thus,the dispersive power of the material is $0.0615$.

The refractive index of a particular material is $1.67$ for blue light,$1.65$ for yellow light,and $1.63$ for red light. The dispersive power of the material is .........

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