what is the value of refractive index of the medium if the critical angle of incidence in a denser-rarer interface is equal to 45 degrees
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Answered by
16
Given:
Angle of incidence =45°
angle of refraction = critical angle=90°
by snell's law,
n1sini=n2 sinr
n1=1 , so n2= ?
n2=sini/sinr
n2=sin45°/sin90°
since sin 90=1
n2= sin 45° =1/√2=0.7071
Angle of incidence =45°
angle of refraction = critical angle=90°
by snell's law,
n1sini=n2 sinr
n1=1 , so n2= ?
n2=sini/sinr
n2=sin45°/sin90°
since sin 90=1
n2= sin 45° =1/√2=0.7071
Answered by
2
Answer:
Correct option is
A
1.41
As we know,
sin45
o= μ1⇒μ=2 =1.41
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