Question #f17c9

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Answer 1 · 2017-01-31T13:03:37

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

$ω = \frac{μ_{b} - μ_{r}}{μ_{y} - 1}$ $omega=(mu_b-mu_r)/(mu_y-1$

where

$μb→refractive index in respect of blue light$ $mu_b->"refractive index in respect of blue light"$

$μr→refractive index in respect of red light$ $mu_r->"refractive index in respect of red light"$

$μy→refractive index in respect of yellow light$ $mu_y->"refractive index in respect of yellow light"$

Given

$ω = 0.031 μ_{b} = 1.665 μ_{r} = 1.645$ $omega=0.031" "mu_b=1.665" "mu_r=1.645$

So inserting the values we have

$ω = \frac{μ_{b} - μ_{r}}{μ_{y} - 1}$ $omega=(mu_b-mu_r)/(mu_y-1$

$\Rightarrow 0.031 = \frac{1.665 - 1.645}{μ_{y} - 1}$ $=>0.031=(1.665-1.645)/(mu_y-1$

$\Rightarrow μ_{y} - 1 = \frac{0.020}{0.031}$ $=>mu_y-1=0.020/0.031$

$\Rightarrow μ_{y} = \frac{20}{31} + 1 = \frac{51}{31} \approx 1.64516$ $=>mu_y=20/31+1=51/31~~1.64516$