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One of the factors that cause a diamond to sparkle is its relatively small critical angle. Compare the critical angle of diamond in air compared to that of diamond in water.

Short Answer

Expert verified
Answer: The critical angle for a diamond in air is approximately 24.4°, whereas the critical angle for a diamond in water is approximately 32.8°. The diamond in air has a smaller critical angle, which means it experiences more total internal reflection, causing it to sparkle more.

Step by step solution

01

First, we need to identify the refractive indices of the materials involved. The refractive index of a diamond is approximately 2.42. The refractive index of air is close to 1, and the refractive index of water is approximately 1.33. #Step 2: Calculate the Critical Angle: Diamond in Air#

The critical angle (theta_c) is the angle of incidence for which the angle of refraction is 90 degrees. To find the critical angle, we can use Snell's Law: \(n_1 \cdot \sin(\theta_c) = n_2 \cdot \sin(90^{\circ})\) Where \(n_1\) is the refractive index of the first medium (diamond), \(n_2\) is the refractive index of the second medium(air), and \(\theta_c\) is the critical angle. In our case, \(n_1 = 2.42\) and \(n_2 = 1\). Solving for \(\theta_c\), we get: \(\theta_c = \arcsin(\frac{n_2}{n_1}) = \arcsin(\frac{1}{2.42})\) #Step 3: Calculate the Critical Angle: Diamond in Water#
02

Now we will calculate the critical angle for a diamond in water. We will use the same formula and replace \(n_2\) with the refractive index of water, which is 1.33. So, we have \(n_1 = 2.42\) and \(n_2 = 1.33\). Solving for \(\theta_c\): \(\theta_c = \arcsin(\frac{n_2}{n_1}) = \arcsin(\frac{1.33}{2.42})\) #Step 4: Compare the Critical Angles#

We can now compare the critical angles for a diamond in air and a diamond in water: Diamond in air: \(\theta_c = \arcsin(\frac{1}{2.42}) \approx 24.4^{\circ}\) Diamond in water: \(\theta_c = \arcsin(\frac{1.33}{2.42}) \approx 32.8^{\circ}\) The critical angle for a diamond in air is smaller than the critical angle for a diamond in water. This means that a diamond will experience more total internal reflection in air, making it sparkle more.

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Key Concepts

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

Refractive Index
The refractive index is a fundamental concept in optics that describes how light propagates through different media. It is denoted by the symbol \( n \) and is defined as the ratio of the speed of light in a vacuum to the speed of light in the medium.

For instance:
  • The refractive index of air is approximately 1, meaning light travels almost as fast in air as it does in a vacuum.
  • Water has a refractive index of about 1.33, showing that light slows down when it enters water.
  • Diamond has a high refractive index of 2.42, which means that light travels much slower in diamond than in air or water.
The higher the refractive index, the more the light will bend, or refract, as it enters the material. This bending of light at the surface is key to understanding phenomena such as the critical angle and total internal reflection.
Critical Angle
The critical angle is an important concept in optics, particularly when discussing the behavior of light at the boundary between two different media.

It is defined as the angle of incidence above which all light is reflected back into the medium rather than being transmitted. To find the critical angle \( \theta_c \), Snell's Law is used:\[ n_1 \cdot \sin(\theta_c) = n_2 \cdot \sin(90^{\circ}) \]Here, \( n_1 \) is the refractive index of the first medium, and \( n_2 \) is the refractive index of the second medium.

For a diamond in air, with air having a refractive index of 1, the critical angle is smaller than for a diamond in water. This results in more light being internally reflected within the diamond when it is in air, contributing to its sparkle.
Total Internal Reflection
Total internal reflection occurs when the angle of incidence exceeds the critical angle. This phenomenon ensures that light is completely reflected back into the original medium, rather than being refracted out.

It is most commonly observed when light travels from a denser material to a less dense one, such as from diamond to air. Since the critical angle for diamond in air is about 24.4 degrees, any angle greater than this will cause total internal reflection, maximizing the light trapped within the diamond.

This maximization enhances the brilliance and sparkle of diamonds when viewed in the air. Total internal reflection is key in fiber optics technology, where it is used to trap light within optical fibers to efficiently transmit data over long distances.
Optics
Optics is a branch of physics focused on studying light and its interactions with matter. It encompasses a variety of phenomena, including refraction, reflection, and dispersion.

Key principles of optics include:
  • Reflection: Light bounces off a surface. This is further categorized into regular and diffuse reflection.
  • Refraction: Light bends when passing through different media. The degree of bending is determined by the refractive index.
  • Dispersion: The separation of light into its component colors, as seen in rainbows.
These principles are used to design lenses, spectacles, microscopes, and other optical devices. Understanding how light behaves at interfaces is crucial for controlling light in technologies ranging from simple glasses to complex lasers.
Diamond Optical Properties
Diamonds are renowned for their optical characteristics, making them highly valuable in both jewelry and industrial applications. Their sparkle is due to several optical properties:

  • High Refractive Index: At 2.42, it causes significant bending of light, contributing to its brilliance.
  • Small Critical Angle: Allows more light to be internally reflected, enhancing the sparkle.
  • Dispersion: Diamonds have a relatively high dispersion, meaning they split white light into its constituent colors, producing a 'fire'.
These optical features not only make diamonds aesthetically appealing but also crucial for applications requiring precision and durability, such as in cutting tools and windows in high-pressure environments.

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Most popular questions from this chapter

An object is located at a distance of \(100 . \mathrm{cm}\) from a concave mirror of focal length \(20.0 \mathrm{~cm}\). Another concave mirror of focal length \(5.00 \mathrm{~cm}\) is located \(20.0 \mathrm{~cm}\) in front of the first concave mirror. The reflecting sides of the two mirrors face each other. What is the location of the final image formed by the two mirrors and the total magnification by the combination?

What is the magnification for a plane mirror? a) +1 c) greater than +1 b) -1 d) not defined for a plane mirror

You are submerged in a swimming pool. What is the maximum angle at which you can see light coming from above the pool surface? That is, what is the angle for total internal reflection from water into air?

Reflection and refraction, like all classical features of light and other electromagnetic waves, are governed by the Maxwell equations. The Maxwell equations are time-reversal invariant, which means that any solution of the equations reversed in time is also a solution. a) Suppose some configuration of electric charge density \(\rho,\) current density \(\vec{j},\) electric field \(\vec{E},\) and magnetic field \(\vec{B}\) is a solution of the Maxwell equations. What is the corresponding time-reversed solution? b) How, then, do "one-way mirrors" work?

Fermat's Principle, from which geometric optics can be derived, states that light travels by a path that minimizes the time of travel between the points. Consider a light beam that travels a horizontal distance \(D\) and a vertical distance \(h\), through two large flat slabs of material, with a vertical interface between the materials. One material has a thickness \(D / 2\) and index of refraction \(n_{1},\) and the second material has a thickness \(D / 2\) and index of refraction \(n_{2} .\) Determine the equation involving the indices of refraction and angles from horizontal that the light makes at the interface \(\left(\theta_{1}\right.\) and \(\theta_{2}\) ) which minimize the time for this travel.

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