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JEE Advanced Model Paper 6 with solutions for online practice
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© examsnet.com
Question : 9
Total: 60
An optical fibre made of glass core of refractive index
µ
g
=
√
3
having an outer covering(called cladding) of refractive index
µ
c
= 1.5 . Total internal reflections will occur at the core– cladding interface if the angle of incidence (i) is less than
30°
45°
60°
75°
Validate
Solution:
The critical angle is given by
sin
i
e
=
µ
e
µ
g
=
1.5
√
3
⇒ sin
i
e
=
√
3
2
⇒
i
e
= 60°
For total internal reflection, the angle of incident at B must be greater than 60ϕ . Angle rwhich is (90° -
i
e
) must be less than 30° i.e.
r
m
a
x
= 30° . Applying Snell’s law at A, we have 1 × sin
i
m
a
x
=
√
3
s
i
n
r
m
a
x
=
√
3
sin30° =
√
3
2
Which gives
i
m
a
x
= 60°
© examsnet.com
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