Question

A beam of white light enters a transparent material. Wavelengths for which the index of refraction is n are refracted at angle ${\theta }_2$.

Wavelengths for which the index of refraction is n + $\delta n$, where $\delta n<<n$, are refracted at angle ${\theta }_2+\delta \theta$.

a). Show that the angular separation of the two wavelengths, in radians, is $\delta \theta =-(\delta n/n)\tan {\theta }_2$.

b). A beam of white light is incident on a piece of glass at 30.0°. Deep violet light is refracted 0.28° more than deep red light. The index of refraction for deep red light is known to be 1.552. What is the index of refraction for deep violet light?

Short answer

$\delta \theta =-(\delta n/n)\tan {\theta }_2$

$\delta n=-(n/\tan {\theta }_2)\delta \theta$

Step-by-step solution

Given values δn=-0.28οnred=1.552θ1=30ο

a). From Snell's law, we have

sin $\theta$₁= n sin $\theta$₂and sin $\theta$₁

=(n+$\delta$n) sin ($\theta$₂+$\delta$$\theta$)

b) we use formula,            δθ=-(δn/n)tanθ2

From previous formula ,we have to find value for$\delta n$