the refractive index of a step index optical fibre is 1.44 and that of a cladding is 1.4 find the value of q critical angle and angle of acceptance
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A silicon optical fibre with a core diameter large enough has a core refractive index of 1.50 and a cladding refractive index 1.47. Determine
(i) the critical angle at the core cladding interface,
(ii) the numerical aperture for the fibre
(iii) the acceptance angle in air for the fibre.
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Solution
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Here,μ
1
=1.50;μ
2
=1.47;
μ
0
=1
(i) Critical angle θ
c
at the core-cladding interface is given by
θ
c
=sin
−1
1.50
1.47
=78.5
0
(ii) Numerical aperture,NA=(μ
1
2
−μ
2
2
)
2
1
=[(1.50)
2
−(1.47)
2
]
2
1
=(2.25−2.16)
1/2
=0.30
(iii) Acceptance angle θ
a
=sin
−1
(NA)
=sin
−1
(0.30)=17.4
0
Explanation:
I'm not sure