If the refractive indices of a flint glass for red, yellow and violet colours are $1.42, 1.44 and 1.48$ respectively then, the dispersive power of the glass is :

0.111

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0.333

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0.136

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0.013

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Solution

The correct option is C $0.136$
Given,
$n_{R} = 1.42$
$n_{Y} = 1.44$
$n_{V} = 1.48$

We know that,
Dispersive power $(ω) = \frac{n_{V} - n_{R}}{n_{Y} - 1}$
So, $ω = \frac{1.48 - 1.42}{1.44 - 1}$
$\Rightarrow ω = \frac{0.06}{0.44}$
$\Rightarrow ω \approx 0.136$

Hence, option $(c)$ is the correct answer.

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If the refractive indices of a flint glass for red, yellow and violet colours are 1.42,1.44 and 1.48 respectively then, the dispersive power of the glass is :

The correct option is C 0.136Given, nR=1.42 nY=1.44 nV=1.48 We know that, Dispersive power (ω)=nV−nRnY−1 So, ω=1.48−1.421.44−1 ⇒ω=0.060.44 ⇒ω≈0.136 Hence, option (c) is the correct answer.

If the refractive indices of a flint glass for red, yellow and violet colours are $1.42, 1.44 and 1.48$ respectively then, the dispersive power of the glass is :