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Optical fibers are constructed with a cylindrical core surrounded by a sheath of cladding material. Common materials used are pure silica \(n_2 = 1.4502\) for the cladding and silica doped with germanium \(n_1 = 1.4652\) for the core. (a) What is the critical angle \(\theta_{crit}\) for light traveling in the core and reflecting at the interface with the cladding material? (b) The numerical aperture (NA) is defined as the angle of incidence \(theta_i\) at the flat end of the cable for which light is incident on the core-cladding interface at angle \(\theta_{crit}\) (\(\textbf{Fig. P33.46}\)). Show that sin \(\theta_i\) =\( \sqrt {n^2_1 - n^2_2}\) . (c) What is the value of \(\theta_i\) for \(n_1\) = 1.465 and \(n_2\) = 1.450?

Short Answer

Expert verified
(a) \( \theta_{crit} \approx 80.57^\circ \); (b) \( \sin(\theta_i) = \sqrt{n_1^2 - n_2^2} \); (c) \( \theta_i \approx 14.02^\circ \).

Step by step solution

01

Understand Refraction and Critical Angle

To find the critical angle \( \theta_{crit} \), we use the concept of total internal reflection that occurs when light moves from a medium with a higher refractive index to one with a lower refractive index. The critical angle is the angle of incidence in the denser medium (core) for which the angle of refraction is 90 degrees in the less dense medium (cladding).
02

Apply Snell's Law

Snell's Law is given by \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \). For the critical angle, we set \( \theta_2 = 90^\circ \), making \( \sin(\theta_2) = 1 \), then the formula becomes \( n_1 \sin(\theta_{crit}) = n_2 \cdot 1 \).
03

Solve for the Critical Angle

Rearrange Snell's Law to solve for \( \theta_{crit} \): \[ \sin(\theta_{crit}) = \frac{n_2}{n_1} \]. Substituting the given values \( n_1 = 1.4652 \) and \( n_2 = 1.4502 \), we have \[ \sin(\theta_{crit}) = \frac{1.4502}{1.4652} \approx 0.9897 \]. Thus, \( \theta_{crit} = \arcsin(0.9897) \approx 80.57^\circ \).
04

Understand Numerical Aperture (NA)

The Numerical Aperture (NA) of an optical fiber describes the range of angles at which the fiber can accept light. It relates to the critical angle by connecting angles of incidence outside and inside the fiber.
05

Derive the Formula for Numerical Aperture

By definition, NA is \( \sin(\theta_i) \) for which light refracts to \( \theta_{crit} \) at the core-cladding interface. Using our understanding of \( n_1 \sin(\theta_{crit}) = n_2 \), the largest \( \theta_i \) occurs when \( \sin(\theta_i) = \sqrt{n_1^2 - n_2^2} \).
06

Calculate the Incidence Angle \( \theta_i \)

Using the formula \( \sin(\theta_i) = \sqrt{n_1^2 - n_2^2} \), we plug in the given \( n_1 = 1.465 \) and \( n_2 = 1.450 \). The calculation yields \[ \sin(\theta_i) = \sqrt{1.465^2 - 1.450^2} \approx 0.242 \]. Therefore, \( \theta_i = \arcsin(0.242) \approx 14.02^\circ \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Total Internal Reflection
Total internal reflection is a fascinating phenomenon that occurs in optical fibers. It happens when light travels from a medium with a higher refractive index, like the fiber core, to a medium with a lower refractive index, such as the cladding.
The light is completely reflected back into the core if the angle of incidence exceeds a certain value, the critical angle. This is important because it allows light to be transmitted over long distances with minimal loss. Optical fibers take advantage of this property to efficiently guide light along their length.
Effects of total internal reflection are extensively used in telecommunications and medical instruments, among other applications. These include transmitting internet signals or creating endoscopes that help doctors look inside the human body.
Critical Angle
The critical angle is the minimum angle of incidence at which light is entirely reflected within a denser medium. When light reaches this angle, it transitions from traveling through the core to being confined within it. This means no refraction occurs beyond the critical angle.
The critical angle depends on the refractive indices of the two materials involved. For the fiber, using Snell's Law, it can be calculated as \( \arcsin\left( \frac{n_2}{n_1}\right) \).
For instance, in an optical fiber where the refractive index of the core \(n_1 = 1.4652\) and the cladding \(n_2 = 1.4502\), the critical angle is about \(80.57^\circ\).
This precise condition helps in ensuring that light continues to propagate within the fiber, which is crucial for efficient signal transmission.
Numerical Aperture
Numerical Aperture (NA) is a measure of how much light can be captured and transmitted through an optical fiber. It indicates the size of the cone of light that enters the fiber.
The NA is based upon the critical angle and is calculated using the indices of refraction of both the core and the cladding. The NA formula is \( \sin(\theta_i) = \sqrt{n_1^2 - n_2^2} \), where \( \theta_i \) is the maximum angle of light entering the fiber that results in total internal reflection.
For example, with core and cladding refractive indices of 1.465 and 1.450, the NA gives us insight into how effectively the fiber can gather light, and we find \(\sin(\theta_i)\) to be approximately 0.242.
This value is vital for ensuring that the fiber captures the maximum amount of light, maintaining strong signal integrity.
Snell's Law
Snell's Law is fundamental in understanding how light travels through different media. It provides the relationship between the angle of incidence and the angle of refraction when light passes from one medium to another.
The formula is \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \( n_1 \) and \( n_2 \) are the refractive indices of the two media, \( \theta_1 \) is the angle of incidence, and \( \theta_2 \) is the angle of refraction.
When applied to finding the critical angle, Snell's Law helps determine the angle beyond which no refraction occurs. This reflects the principle of total internal reflection, a vital aspect of optical fiber technology.
Understanding Snell's Law allows us to predict and manipulate how light behaves when entering different materials, which is essential for designing efficient fiber optics systems.

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Most popular questions from this chapter

A beam of light has a wavelength of 650 nm in vacuum. (a) What is the speed of this light in a liquid whose index of refraction at this wavelength is 1.47? (b) What is the wavelength of these waves in the liquid?

The critical angle for total internal reflection at a liquid air interface is 42.5\(^\circ\). (a) If a ray of light traveling in the liquid has an angle of incidence at the interface of 35.0\(^\circ\), what angle does the refracted ray in the air make with the normal? (b) If a ray of light traveling in air has an angle of incidence at the interface of 35.0\(^\circ\), what angle does the refracted ray in the liquid make with the normal?

A beam of light is traveling inside a solid glass cube that has index of refraction 1.62. It strikes the surface of the cube from the inside. (a) If the cube is in air, at what minimum angle with the normal inside the glass will this light \(not\) enter the air at this surface? (b) What would be the minimum angle in part (a) if the cube were immersed in water?

A ray of light traveling in water is incident on an interface with a flat piece of glass. The wavelength of the light in the water is 726 nm, and its wavelength in the glass is 544 nm. If the ray in water makes an angle of 56.0\(^\circ\) with respect to the normal to the interface, what angle does the refracted ray in the glass make with respect to the normal?

The refractive index of a certain glass is 1.66. For what incident angle is light reflected from the surface of this glass completely polarized if the glass is immersed in (a) air and (b) water?

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