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Chapter 23: Problem 30

What is the index of refraction of the core of an optical fiber if the cladding has \(n=1.20\) and the critical angle at the core-cladding boundary is \(45.0^{\circ} ?\)

Short Answer

Expert verified
In summary, given the index of refraction of the cladding and the critical angle at the core-cladding boundary, we used Snell's law and determined that the index of refraction of the core of the optical fiber is approximately 1.70.

Step by step solution

01

Recall the definition of critical angle

The critical angle is the angle of incidence in the denser medium (core) for which the angle of refraction in the less dense medium (cladding) becomes \(90^{\circ}\).
02

Use Snell's Law

Snell's law relates the angles of incidence and refraction with the indices of refraction of the two media. The formula for Snell's law is: \(n_{1} \sin \theta_{1} = n_{2} \sin \theta_{2}\) In our case, \(\theta_{1}\) is the critical angle in the core, \(\theta_{2} = 90^{\circ}\), \(n_{1}\) is the index of refraction of the core, and \(n_{2}\) is the index of refraction of the cladding.
03

Solve for the index of refraction of the core

Since we need to find the index of refraction of the core (\(n_{1}\)), we can rearrange Snell's law to solve for \(n_{1}\): \(n_{1} = \frac{n_{2} \sin \theta_{2}}{\sin \theta_{1}} \) We are given \(n_{2} = 1.20\) and \(\theta_{1} = 45.0^{\circ}\). Since \(\theta_{2} = 90^{\circ}\), \(\sin \theta_{2} = 1\). Thus, our formula becomes: \(n_{1} = \frac{1.20 \cdot 1}{\sin 45.0^{\circ}}\)
04

Calculate the value of \(n_1\)

Now, we can plug in the values and find the index of refraction \(n_1\) of the core: \(n_{1} = \frac{1.20}{\sin 45.0^{\circ}}\) \(n_{1} = \frac{1.20}{0.7071}\) (using the sine value of the 45-degree angle) \(n_{1} \approx 1.70\) So, the index of refraction of the core of the optical fiber is approximately 1.70.

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Most popular questions from this chapter

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