If the refractive index of glass is 1.5, what is the critical angle at the air-glass interface?
a
sin-1\(\frac{1}{2}\)
b
sin-1\(\frac{2}{3}\)
c
sin-1\(\frac{3}{4}\)
d
sin-1\(\frac{8}{9}\)
Explanation
Correct Option
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Redhawk
2 years ago
c is critical incidence while n is refractive index
sin c = 1/n
so n=1.5 which is 3/2 in fraction
import it in the equation
sin c = 1÷3/2
sin c = 1* 2/3
sin c =2/3
then
c = sin-¹ 2/3
hope this helps
