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Chapter 20: Problem 2756

A silicon optical fiber has a core refractive index of \(1.50\) and a cladding refractive index \(1.47\) what is the numerical aperture of the fiber? (A) \(0.90\) (B) \(0.60\) (C) \(0.45\) (D) \(0.30\)

Short Answer

Expert verified
The numerical aperture of the silicon optical fiber with core refractive index 1.50 and cladding refractive index 1.47 is approximately \(0.30\).

Step by step solution

01

Write down the given values

We are given the core refractive index (n1) and the cladding refractive index (n2) of a silicon optical fiber. \[n1 = 1.50\] \[n2 = 1.47\]
02

Calculate the squared difference between n1 and n2

We first need to find the squared difference between the core refractive index and the cladding refractive index. \[(n1^2 - n2^2) = (1.50^2 - 1.47^2) = (2.25 - 2.1609)\]
03

Calculate the numerical aperture (NA)

Now, apply the formula for the numerical aperture (NA) \[NA = \sqrt{n1^2 - n2^2} \] Use the squared difference calculated in Step 2: \[NA = \sqrt{2.25 - 2.1609} = \sqrt{0.0891}\] Now, we find the square root of 0.0891: \[NA \approx 0.2985\]
04

Choose the correct answer

The calculated numerical aperture (NA) is approximately 0.2985, which is closest to the given option (D) \(0.30\). Hence, the correct answer is: (D) \(0.30\)

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