Questions
Relative permittivity of a medium is 4 and critical angle of it is 45°. Then its relative permeability is
a
22
b
42
c
0.5
d
2
detailed solution
Correct option is C
velocity of light in free space is c=1μ0ε0velocity of light in a medium is v=1μεrefractive index n=cv=μεμ0ε0= μrεr=4εr sin C=1n C is 450 n=22=4εr εr=0.5Talk to our academic expert!
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