Question #4c1f6

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Answer 1 · 2017-01-31T13:23:33

We know the expression for dispersive power $omega$ as follows

$omega=(mu_F-mu_C)/(mu_D-1$

where

$mu_F->"refractive index in respect of F light"$

$mu_C->"refractive index in respect of C light"$

$mu_D->"refractive index in respect of D light"$

Given

$mu_F=1.805" "mu_C=1.790" "mu_D=$

So inserting the values we have

$omega=(mu_F-mu_C)/(mu_D-1$

$=>omega=(1.805-1.790)/(1.795-1)=0.015/0.795~~0.0188$