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 :

0.031

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1.60

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0.0615

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0.024

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Solution

The correct option is C $0.0615$
Dispersive power of material= $\frac{n_{b} - n_{r}}{n_{y} - 1}$
$= \frac{1.67 - 1.63}{1.65 - 1} = 0.0615$

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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 :

The correct option is C 0.0615Dispersive power of material=nb−nrny−1=1.67−1.631.65−1=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 :

The correct option is C $0.0615$
Dispersive power of material= $\frac{n_{b} - n_{r}}{n_{y} - 1}$
$= \frac{1.67 - 1.63}{1.65 - 1} = 0.0615$