Question

PASSAGE A glass clad fiber is made with core glass of refractive index 1.4 and the cladding is doped to give a fractional refractive index difference of 0.004.
1. the acceptance angle of the cable
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Answer 1

Given:

The class of refractive index $1.4$ and the cladding doped index difference of $0.004$

To find:

The acceptance angle of the cable

Explanation:

Let the refractive index of cladding be $n_{2}$, so we have,

Δ$=\frac{n_{1}-n_{2}}{n_{1}}$

$0.004=\frac{1.4-n_{2}}{1.4}$

$n_{2}=1.4-1.4*0.004$

=$1.3944$

Answer:

Therefore, The acceptance angle of the cable in the $1.3944$.