/*! 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 62 Consider the common mirage assoc... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the common mirage associated with an inhomogeneous distribution of air situated above a warm roadway. Envision the bending of the rays as if it were instead a problem in total internal reflection. If an observer, at whose head \(n_{a}=1.00029,\) sees an apparent wet spot at \(\theta_{t} \geq 88.7^{\circ}\) down the road, find the index of the air immediately above the road.

Short Answer

Expert verified
The index of refraction of the air above the road is approximately 1.0013.

Step by step solution

01

Understanding Total Internal Reflection

Total internal reflection occurs when light passing through a medium with a higher index of refraction cannot pass into a medium with a lower index of refraction. Instead, it reflects entirely within the medium. This happens when the angle of incidence inside the denser medium exceeds the critical angle.
02

Snell's Law for Mirage Problem

In this problem, we use Snell's Law: \( n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \), where \( n_1 \) is the refractive index of the denser medium (roadside air), \( n_2 \) the less dense medium (air above the road), and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively. Since light reflects internally, assume \( \theta_2 = 90^{\circ} \), where all the light reflects back.
03

Determine Critical Angle for Total Internal Reflection

To find the critical angle \( \theta_c \) from Snell's Law, set \( \theta_2 = 90^{\circ} \) (as it approaches total internal reflection). Then, the equation simplifies to \( \sin \theta_c = \frac{n_2}{n_1} \). Given \( n_2 = 1.00029 \) (air) and \( \theta_t = \theta_c = 88.7^{\circ} \).
04

Solving for the Index of Refraction

Rearrange Snell's Law to solve for \( n_1 \): \( n_1 = \frac{n_2}{\sin \theta_c} \). Substitute \( n_2 = 1.00029 \) and \( \theta_c = 88.7^{\circ} \). Thus: \( n_1 = \frac{1.00029}{\sin(88.7^{\circ})} \). Use \( \sin(88.7^{\circ}) \approx 0.999 \) which gives the approximate calculations.
05

Calculate the Index of Refraction

Continue with the calculation: \( n_1 = \frac{1.00029}{0.999} \approx 1.0013 \). This result gives the refractive index of the air layer immediately above the warm road surface.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

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 optical phenomenon where light reflects entirely within a medium, without transmitting to a second medium. This occurs when light travels from a dense medium to a less dense medium, such as from water to air. The angle of incidence must be greater than a certain threshold, known as the critical angle, for total internal reflection to occur.

Light behaves like this because at larger angles, the light cannot bend enough to exit into the less dense medium; instead, it reflects back as if hitting a mirror. This is why we see effects like sparkling in diamonds, fiber optics effectively transmitting data, or conditions for a mirage along hot roadways.
Snell's Law
Snell's Law is fundamental in understanding how light behaves when passing through different media. It describes how light bends, or refracts, as it moves from one medium into another with a different refractive index. The law is mathematically expressed as \[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]

Where:
  • \( n_1 \) and \( n_2 \) are the refractive indices of the two media.
  • \( \theta_1 \) is the angle of incidence in the first medium.
  • \( \theta_2 \) is the angle of refraction in the second medium.
The equation shows how the sine of the angles is directly proportional to the refractive indices, which determines the direction and extent of bending. Understanding Snell's Law helps explain phenomena such as the apparent bending of a straw in water or the dynamics of mirages over hot surfaces.
Refraction
Refraction is when light changes direction as it passes from one medium into another. This is due to the change in speed of light in different materials. For instance, light travels slower in glass than in air, which causes the light path to bend or angle at an interface, such as glass to air.

This bending effect is why objects appear misaligned at an interface; a common illustration is a straight stick that looks bent when partially submerged in water. Refraction is crucial in optics, affecting everything from camera lenses collecting light to human eyes focusing images.

Without refraction, we wouldn't have the focus required for vision or the instrumental devices like microscopes and telescopes that aid scientific discovery.
Mirage
A mirage is an interesting optical illusion that occurs commonly on hot days. It happens when layers of air at different temperatures cause light to refract in unusual ways. Typically, a warm road heats the air above it, making the air layers less dense closer to the surface.

As light travels from the sky and enters these varying air layers, it bends steadily until it reaches a point of total internal reflection, then deflects towards the observer's eye. The light's rapid directional change creates the illusion of water or a reflective surface on the road, explaining the shimmering images seen in hot conditions.
Critical Angle
The concept of a critical angle is vital for understanding total internal reflection. It is the minimum angle of incidence at which light can strike the interface between two media and be completely reflected back into the denser medium.

Mathematically, the critical angle \( \theta_c \) is determined from Snell's Law by setting the angle of refraction \( \theta_2 \) to 90 degrees, meaning the refracted ray grazes along the interface. The formula is:\[ \sin(\theta_c) = \frac{n_2}{n_1} \]

where \( n_1 \) is the refractive index of the denser medium and \( n_2 \) is that of the less dense medium. Knowing the critical angle helps in optimizing light paths in technologies like fiber optics and in understanding naturally occurring phenomena like mirages.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The equation for a driven damped oscillator is $$m_{e} \ddot{x}+m_{e} \gamma \dot{x}+m_{e} \omega_{0}^{2} x=q_{e} E(t)$$ (a) Explain the significance of each term. (b) Let \(E=E_{0} e^{i \omega t}\) and \(x=x_{0} e^{i(\omega t-\alpha)},\) where \(E_{0}\) and \(x_{0}\) are real quan tities. Substitute into the above expression and show that $$x_{0}=\frac{q_{e} E_{0}}{m_{e}} \frac{1}{\left[\left(\omega_{0}^{2}-\omega^{2}\right)^{2}+\gamma^{2} \omega^{2}\right]^{1 / 2}}$$ (c) Derive an expression for the phase lag, \(\alpha,\) and discuss how \(\alpha\) varies as \(\omega\) goes from \(\omega \ll \omega_{0}\) to \(\omega=\omega_{0}\) to \(\omega \gg \omega_{0}\).

A white floodlight beam crosses a large volume containing a tenuous molecular gas mixture of mostly oxygen and nitrogen. Compare the relative amount of scattering occurring for the yellow (580 \(\mathrm{nm}\) component with that of the violet \((400 \mathrm{nm})\) component.

A beam of light in air strikes the surface of a smooth piece of plastic having an index of refraction of 1.55 at an angle with the normal of \(20.0^{\circ} .\) The incident light has component \(E\) -field amplitudes parallel and perpendicular to the plane-of-incidence of \(10.0 \mathrm{V} / \mathrm{m}\) and \(20.0 \mathrm{V} / \mathrm{m},\) respectively. Determine the corresponding reflected field amplitudes.

Using a block of a transparent, unknown material, it is found that a beam of light inside the material is totally internally reflected at the air-block interface at an angle of \(48.0^{\circ} .\) What is its index of refraction?

Compare the amplitude reflection coefticicnts lor an inr-water \(\left(n_{\mathrm{u}}=4 / 3\right)\) interface with that of an arr-ciown glass \(\left(n_{\mathrm{c}}=3 / 2\right)\) inicer face, both at near-normal incidence. What are the contesponche ratios of the reflected to the incident intadiances?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.