Question #106374
Find the critical angle for light traveling from a ruby (n=1.776) into air
1
Expert's answer
2020-03-25T11:01:28-0400

Critical angle from ruby in air


sin(α)=1nsin(\alpha)=\frac{1}{n}α=arcsin(1n)=arcsin(11.776)=34o\alpha=arcsin(\frac{1}{n})=arcsin(\frac{1}{1.776})=34^o


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