/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 Diamond has an index of refracti... [FREE SOLUTION] | 91Ó°ÊÓ

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Diamond has an index of refraction of \(2.42\), and part of the reason diamonds sparkle is that this encourages a light ray to undergo many total internal reflections before it emerges. (a) Calculate the critical angle at which total internal reflection occurs in diamond. (answer check available at lightandmatter.com) (b) Explain the interpretation of your result: Is it measured from the normal, or from the surface? Is it a minimum, or a maximum? How would the critical angle have been different for a substance such as glass or plastic, with a lower index of refraction?

Short Answer

Expert verified
(a) The critical angle is approximately \(24.4^\circ\). (b) It's measured from the normal, is a minimum threshold, and would be larger for materials with lower refractive indices.

Step by step solution

01

Understand Total Internal Reflection

Total internal reflection occurs when a light ray traveling from a denser medium to a less dense one hits the boundary at an angle greater than the critical angle, causing it to reflect back into the original medium. This phenomenon requires an understanding of the concept of critical angle, which depends on the index of refraction of the media involved.
02

Recall Snell's Law and Critical Angle Formula

Snell's Law states that for two media with refractive indices \( n_1 \) and \( n_2 \), and angles of incidence \( \theta_1 \) and refraction \( \theta_2 \), it holds that \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \). The critical angle \( \theta_c \) is given by \( \sin(\theta_c) = \frac{n_2}{n_1} \) where \( n_2 < n_1 \). In this problem, \( n_1 = 2.42 \) for diamond and \( n_2 \approx 1 \) for air.
03

Solve for the Critical Angle

Plug in the values into the critical angle formula: \( \sin(\theta_c) = \frac{1}{2.42} \). Calculate to find \( \theta_c \): \[ \theta_c = \arcsin\left(\frac{1}{2.42}\right) \approx 24.4^\circ \].
04

Interpret the Critical Angle

The critical angle is measured from the normal to the surface; thus, it represents the minimum angle of incidence at which total internal reflection occurs. This angle is a threshold value that is specific to the medium of diamond. If the medium had a lower index of refraction, like glass or plastic, the critical angle would be larger, allowing more angles to cause total internal reflection.

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Key Concepts

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

Critical Angle
The critical angle is a fascinating concept that unlocks the magic behind total internal reflection. When light travels through a medium like diamond towards a less dense medium like air, it encounters a boundary. If the angle of incidence is larger than this critical angle, light doesn't pass into the air but rather bounces back into the diamond.
Think of the critical angle as a special threshold. It is the minimum angle at which this reflection happens. For diamond, which has a high index of refraction of 2.42, the critical angle is relatively small, around 24.4 degrees. This means that it doesn't take much tilt for light to stay within the diamond, contributing to its sparkle.
Remember, the critical angle is always measured from the normal. Picture a line perpendicular to the surface at the point of incidence. The critical angle is the angle between the incoming light ray and this line. Different materials will have different critical angles:
  • Materials with a high index of refraction, like diamond, have a small critical angle.
  • Materials with a lower index, such as glass or plastic, will have a larger critical angle. This means light will more easily pass into the less dense medium, reflecting less inside them.
Understanding the critical angle helps us grasp why materials like diamonds are so dazzling.
Snell's Law
Snell's Law is a key principle that explains how light behaves at the boundary between two different media. This law is like a rulebook for light, determining how it bends, or refracts, when moving from one material to another.
According to Snell's Law, the relationship between the angle of incidence (the angle at which a light ray hits the surface) and the angle of refraction (the angle at which light travels in the new medium) is determined by the indices of refraction of the two media involved. The mathematical formula is:
\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]
Where:
  • \( n_1 \) and \( n_2 \) are the indices of refraction for the original and new media, respectively.
  • \( \theta_1 \) is the angle of incidence.
  • \( \theta_2 \) is the angle of refraction.
In our diamond example, the refractive index of diamond (\( n_1 = 2.42 \)) and air (\( n_2 \approx 1 \)) are used to determine the path of light. When conditions are right, Snell's Law also helps in calculating the critical angle, where total internal reflection begins. By setting the angle of refraction to 90 degrees (light skimming along the boundary), the formula adjusts itself into the critical angle equation:
\[ \sin(\theta_c) = \frac{n_2}{n_1} \]
Creating a bridge between refraction and reflection, Snell's Law provides a clear understanding of how and why light bends and reflects in different media.
Index of Refraction
The index of refraction is a valuable property that characterizes how much a particular medium can bend light. Every transparent material has an index of refraction, denoted by \( n \), which is a measure of how much slower light travels in the material compared to in vacuum.
In practical terms, the index of refraction determines both how light bends when entering a new medium and the critical angle for total internal reflection. Here are some key points:
  • The higher the index, the slower light travels through the medium, and the more it bends.
  • Materials like diamond, with a high index (2.42), bend light significantly, making them sparkle as light reflects internally multiple times before exiting.
  • Conversely, materials like air, which have a low index of approximately 1, cause minimal bending/refracting of light.
The concept of refraction index helps us predict and understand not just everyday occurrences of light bending in substances like water or glass, but also exquisite phenomena like the brilliance of diamonds. By observing how light behaves, designers can craft jewelry that enhances reflection and sparkle, maximizing the beauty derived from their refractive properties.

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Most popular questions from this chapter

It would be annoying if your eyeglasses produced a magnified or reduced image. Prove that when the eye is very close to a lens, and the lens produces a virtual image, the angular magnification is always approximately equal to 1 (regardless of whether the lens is diverging or converging).

A converging mirror has focal length \(f\). An object is located at a distance \((1+\epsilon) f\) from the mirror, where \(\epsilon\) is small. Find the distance of the image from the mirror, simplifying your result as much as possible by using the assumption that \(\epsilon\) is small.

When you take pictures with a camera, the distance between the lens and the film has to be adjusted, depending on the distance at which you want to focus. This is done by moving the lens. If you want to change your focus so that you can take a picture of something farther away, which way do you have to move the lens? Explain using ray diagrams.

When swimming underwater, why is your vision made much clearer by wearing goggles with flat pieces of glass that trap air behind them? [Hint: You can simplify your reasoning by considering the special case where you are looking at an object far away, and along the optic axis of the eye.]

Based on Snell's law, explain why rays of light passing through the edges of a converging lens are bent more than rays passing through parts closer to the center. It might seem like it should be the other way around, since the rays at the edge pass through less glass - - - shouldn't they be affected less? In your answer: \- Include a ray diagram showing a huge, full-page, close-up view of the relevant part of the lens. \- Make use of the fact that the front and back surfaces aren't always parallel; a lens in which the front and back surfaces are always parallel doesn't focus light at all, so if your explanation doesn't make use of this fact, your argument must be incorrect. \- Make sure your argument still works even if the rays don't come in parallel to the axis.

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