Question #273774
  1. The speed of light in a material is 0.52c. What is the critical angle of a light ray at the interface between the material and a vacuum?
1
Expert's answer
2021-12-01T22:46:55-0500

voflightinMaterial=0.52cθc=sin1(nlightnmaterial)nmaterial=cvnmaterial=c0.52cnmaterial=10.52nmaterial=1.92θc=sin1(11.92)θc=31.39°θc=31°{v\,of\,light\,in\,Material=0.52c}\\ {\theta_c=sin^{-1}(\frac{n_{light}}{n_{material}})}\\ {n_{material}=\frac{c}{v}}\\ {n_{material}=\frac{c}{0.52c}}\\ {n_{material}=\frac{1}{0.52}}\\ {n_{material}=1.92}\\ {\theta_c=sin^{-1}(\frac{1}{1.92})}\\ {\theta_c=31.39\degree}\\ {\theta_c=31\degree}\\


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS