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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. \(\textbf{P33.51}\). 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 \(\delta\) is given by $${ \delta = \mathrm{arcsin} ({{nR}\over{R + h}}}) - \mathrm{arcsin}({{{R}\over R + h}}) $$ where \(R\) = 6378 km is the radius of the earth. (b) Calculate \(\delta\) using \(n\) = 1.0003 and \(h\) = 20 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.)

Short Answer

Expert verified
\( \delta \approx 0.297^\circ \), which is slightly larger than the sun's angular radius of \(0.25^\circ\).

Step by step solution

01

Understanding the Problem

We need to find the apparent displacement angle \( \delta \) caused by refraction of sunlight in the Earth's atmosphere, assuming the atmosphere has a uniform refraction index of \( n \). We will begin by deriving the formula for \( \delta \) using given values and requirements.
02

Derive the Refraction Formula

Given a light ray entering the atmosphere at the horizon: use Snell's Law at the boundary between space and the atmosphere. Assume \( R \) is the Earth's radius and \( h \) is the atmosphere's height. Snell's Law: \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), where \( n_1 = 1 \), \( \theta_1 = 90^\circ \) - the angle corresponding to the vertical. Transform this into the relationship between \( \delta \) and the refracted angle: \({ \sin \theta_2 = \frac{R}{R + h}} \). With boundary conditions: \( n \sin 0 = \sin(\theta_1) \) and \( \sin(\theta_2) = \frac{R}{R+h} \), rearrange it to solve for \( \theta_1 \) and \( \delta \) using \( \delta = \theta_2 - \theta_1 \).
03

Substitute Values into the Formula

Use \( n = 1.0003 \), \( R = 6378 \text{ km} \), and \( h = 20 \text{ km} \). Calculate \( \theta_2 = \arcsin(\frac{6378}{6398}) \) and \( \theta_1 = \arcsin(\frac{1.0003 \times 6378}{6398}) \). Find \( \delta = \theta_1 - \theta_2 \).
04

Compute Angle \( \delta \)

The calculation using arcsin() functions and the provided values yield \( \delta = \arcsin(1.0003 \times \frac{6378}{6398}) - \arcsin(\frac{6378}{6398}) \). Numerically solve using a calculator to find \( \delta \).
05

Compare with the Sun's Angular Radius

By calculating \( \delta \), compare it with the angular radius of the sun, which is about \(0.25^\circ\). This helps to understand the visual impact of atmospheric refraction on the apparent position of the sun.

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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 in optics that describes how light bends when it passes from one medium to another. It is defined by the equation \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), where \( n_1 \) and \( n_2 \) are the indices of refraction for the two media, and \( \theta_1 \) and \( \theta_2 \) are the angles of incidence and refraction, respectively.
In the context of atmospheric refraction, Snell's Law explains how sunlight refracts as it enters the Earth's atmosphere from space, which generally has a lower refractive index compared to the atmosphere. Since space has an approximate refractive index of 1, and the Earth's atmosphere is slightly higher at about 1.0003, the sunlight bends towards the normal as it enters the atmosphere.
This bending of light changes the apparent position of celestial objects, such as the sun, making them appear higher in the sky than they actually are. Snell's Law becomes crucial in calculating the precise degree of displacement due to this refractive effect.
Angle of Displacement
The angle of displacement, often denoted by \( \delta \), refers to the apparent shift in the position of an object, in this case, the sun, due to the refraction of light. As the sun's rays enter the Earth's atmosphere, they bend towards the Earth's surface because of the higher refractive index of the atmosphere compared to outer space.
To derive the angle \( \delta \), we use the refraction formula \( \delta = \arcsin\left(\frac{nR}{R + h}\right) - \arcsin\left(\frac{R}{R + h}\right) \). Here, \( R \) is the radius of the Earth, \( h \) is the height of the atmosphere, and \( n \) is the uniform index of refraction. This formula allows us to compute \( \delta \) by determining the difference between the apparent angle \( \theta_1 \) and the refracted angle \( \theta_2 \).
The angle of displacement is essential to understanding how refraction affects what we see from the Earth's surface. It explains why the sun and stars appear slightly above the horizon even when they are geometrically below it.
Uniform Index of Refraction
The index of refraction quantifies how much a medium, such as air or water, bends light passing through it. A uniform index of refraction implies that this bending property remains constant throughout the medium.
In the simplified model of the Earth's atmosphere presented in the exercise, the atmosphere has a uniform index of refraction, denoted as \( n = 1.0003 \). This means that the entire atmosphere bends light by the same amount, which simplifies calculations by avoiding changes due to varying densities.
Although in reality, the Earth's atmosphere has a gradually changing refractive index because of varying temperature, pressure, and composition, assuming a uniform index helps focus on foundational concepts without complicating the mathematics. The uniform index model lets us predict the apparent displacement of the sun from a simpler perspective, useful for introductory educational purposes.

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Most popular questions from this chapter

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?

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?

Physicians use high-frequency (\(f\) = 1\(-\)5 MHz) sound waves, called ultrasound, to image internal organs. The speed of these ultrasound waves is 1480 m\(/\)s in muscle and 344 m\(/\)s in air. We define the index of refraction of a material for sound waves to be the ratio of the speed of sound in air to the speed of sound in the material. Snell's law then applies to the refraction of sound waves. (a) At what angle from the normal does an ultrasound beam enter the heart if it leaves the lungs at an angle of 9.73\(^\circ\) from the normal to the heart wall? (Assume that the speed of sound in the lungs is 344 m\(/\)s.) (b) What is the critical angle for sound waves in air incident on muscle?

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?

A flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass is incident on the interface with the water at an angle of \(\theta_a = 36.2{^\circ}\), the ray refracted into the water makes an angle of 49.8\(^\circ\) with the normal to the interface. What is the smallest value of the incident angle \(\theta_a\) for which none of the ray refracts into the water?

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