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Chapter 33: Problem 23

A ray of light in diamond (index of refraction 2.42 ) is incident on an interface with air. What is the largest angle the ray can make with the normal and not be totally reflected back into the diamond?

Short Answer

Expert verified
The largest angle is approximately 24.4 degrees.

Step by step solution

01

Understanding the Concept

To solve this problem, we need to find the critical angle at which light traveling from a medium with a higher index of refraction (diamond) to a medium with a lower index of refraction (air) is no longer transmitted but instead undergoes total internal reflection. The critical angle can be calculated using Snell's Law.
02

Snell's Law Formula

Snell's Law is given by the equation: \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \( n_1 \) and \( n_2 \) are the refractive indices of the two media, and \( \theta_1 \) and \( \theta_2 \) are the angles with respect to the normal. For the critical angle \( \theta_c \), \( \theta_2 = 90^{\circ} \) and \( \sin(\theta_2) = 1 \).
03

Critical Angle Formula

Substituting into Snell's law for \( \theta_2 = 90^{\circ} \), we have: \( n_{diamond} \sin(\theta_c) = n_{air} \cdot 1 \). Therefore, the critical angle \( \theta_c \) can be calculated as \( \sin(\theta_c) = \frac{n_{air}}{n_{diamond}} \).
04

Substitute Refractive Indices

The refractive index of diamond is given as 2.42, and the refractive index of air is approximately 1. Substituting these values into the critical angle formula gives \( \sin(\theta_c) = \frac{1}{2.42} \).
05

Calculate the Critical Angle

Calculate \( \theta_c \) by taking the inverse sine (arcsin) of \( \frac{1}{2.42} \): \( \theta_c = \sin^{-1}(\frac{1}{2.42}) \). Using a calculator, we find \( \theta_c \approx 24.4^{\circ} \).
06

Conclusion

Therefore, the largest angle the ray can make with the normal, and not be totally reflected back into the diamond, is approximately \( 24.4^{\circ} \).

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Key Concepts

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

Snell's Law
Snell's Law is a fundamental principle used to describe how light behaves when it moves between different media. It explains the change in speed and direction of light as it travels through substances with different refractive indices. The law is mathematically expressed as:
  • \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \)
Here, \( n_1 \) and \( n_2 \) are the refractive indices of the two materials through which light passes. \( \theta_1 \) is the angle of incidence, and \( \theta_2 \) is the angle of refraction. A key application of Snell's Law is determining the critical angle, which is the specific angle of incidence beyond which total internal reflection occurs when light travels from a denser medium to a less dense medium.
Total Internal Reflection
Total internal reflection is a fascinating phenomenon that occurs when light attempts to move from a dense medium, like diamond, to a less dense medium, such as air. When the angle of incidence exceeds the critical angle, rather than passing into the second medium, the light is completely reflected back into the first medium.
This effect is not just theoretical; it is used in practical applications such as fiber optics. Fiber optic cables rely on total internal reflection to transmit data over long distances with minimal loss of signal.
  • Occurs when light hits the interface at an angle greater than the critical angle.
  • Results in the light being reflected rather than refracted.
Refractive Index
The refractive index is a measure of how much light slows down as it travels through a particular medium. It is represented by the symbol \( n \), and calculated by comparing the speed of light in a vacuum to the speed of light in the medium. For instance, diamond has a high refractive index of 2.42, indicating that light significantly slows down as it passes through it.
This explains why objects under water appear bent or distorted. The higher the refractive index of a material, the more it bends light away from its original path.
  • A high refractive index means light bends more.
  • Each material has its own unique refractive index.
Optics
Optics is the branch of physics that studies light and its interactions with different substances. It covers a broad range of phenomena, including reflection, refraction, and the behavior of lenses and mirrors. By understanding the principles of optics, such as Snell's Law, scientists and engineers design various optical instruments and technologies.
Common devices that rely on optics include:
  • Telescopes for observing distant astronomical objects.
  • Microscopes for magnifying tiny details.
  • Cameras for capturing images.
Each of these tools manipulates light in specific ways to serve their unique purposes.
Light Refraction
Light refraction occurs when a wave of light changes direction as it moves from one medium into another where its speed is different. This bending happens because of the change in wave velocity as dictated by the medium's refractive index.
Everyday examples of refraction are quite common and include:
  • Prisms dispersing light into a spectrum.
  • A straw appearing bent when placed in a glass of water.
Understanding refraction is essential for comprehending how lenses focus light, which is crucial for correcting vision or designing photographic equipment.

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Most popular questions from this chapter

A parallel beam of light in air makes an angle of \(47.5^{\circ}\) with the surface of a glass plate having a refractive index of 1.66 . (a) What is the angle between the reflected part of the beam and the surface of the glass? (b) What is the angle between the refracted beam and the surface of the glass?

A quarter-wave plate converts linearly polarized light to circularly polarized light. Prove that a quarter-wave plate also converts circularly polarized light to linearly polarized light.

When the sun is either rising or setting and appears to be just on the horizon, it is in fact below the horizon. The explanation for this seeming paradox is that light from the sun bends slightly when entering the earth's atmosphere, as shown in Fig. 33.53 . Since our perception is based on the idea that light travels in straight lines, we perceive the light to be coming from an apparent position that is an angle \(\delta\) above the sun's true position. (a) Make the simplifying assumptions that the atmosphere has uniform density, and hence uniform index of refraction \(n\) , and extends to a height \(h\) above the earth's surface, at which point it abruptly stops. Show that the angle 8 is given by $$ \delta=\arcsin \left(\frac{n R}{R+h}\right)-\arcsin \left(\frac{R}{R+h}\right) $$ where \(R=6378 \mathrm{km}\) is the radius of the earth. (b) Calculate \(\delta\) using \(n=1.0003\) and \(h=20 \mathrm{km}\) . How does this compare to the angular radius of the sun, which is about one quarter of a degree? (In actuality a light ray from the sun bends gradually, not abruptly, since the density and refractive index of the atmosphere change gradually with altitude.)

A beaker with a mirrored bottom is filled with a liquid whose index of refraction is \(1.63 .\) A light beam strikes the top surface of the liquid at an angle of \(42.5^{\circ}\) from the normal. At what angle from the normal will the beam exit from the liquid after traveling down through the liquid, reflecting from the mirrored bottom, and returning to the surface?

Prove that a ray of light reflected from a plane mirror rotates through an angle of 2\(\theta\) when the mirror rotates through an angle \(\theta\) about an axis perpendicular to the plane of incidence.
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