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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. 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 δ 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 δ is given by δ=arcsin(nRR+h)arcsin(RR+h) where R = 6378 km is the radius of the earth. (b) Calculate δ 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
δ0.297, which is slightly larger than the sun's angular radius of 0.25.

Step by step solution

01

Understanding the Problem

We need to find the apparent displacement angle δ 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 δ 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: n1sinθ1=n2sinθ2, where n1=1, θ1=90 - the angle corresponding to the vertical. Transform this into the relationship between δ and the refracted angle: sinθ2=RR+h. With boundary conditions: nsin0=sin(θ1) and sin(θ2)=RR+h, rearrange it to solve for θ1 and δ using δ=θ2θ1.
03

Substitute Values into the Formula

Use n=1.0003, R=6378 km, and h=20 km. Calculate θ2=arcsin(63786398) and θ1=arcsin(1.0003×63786398). Find δ=θ1θ2.
04

Compute Angle δ

The calculation using arcsin() functions and the provided values yield δ=arcsin(1.0003×63786398)arcsin(63786398). Numerically solve using a calculator to find δ.
05

Compare with the Sun's Angular Radius

By calculating δ, compare it with the angular radius of the sun, which is about 0.25. 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 n1sinθ1=n2sinθ2, where n1 and n2 are the indices of refraction for the two media, and θ1 and θ2 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 δ, 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 δ, we use the refraction formula δ=arcsin(nRR+h)arcsin(RR+h). 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 δ by determining the difference between the apparent angle θ1 and the refracted angle θ2.
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 parallel beam of unpolarized light in air is incident at an angle of 54.5 (with respect to the normal) on a plane glass surface. The reflected beam is completely linearly polarized. (a) What is the refractive index of the glass? (b) What is the angle of refraction of the transmitted beam?

Given small samples of three liquids, you are asked to determine their refractive indexes. However, you do not have enough of each liquid to measure the angle of refraction for light refracting from air into the liquid. Instead, for each liquid, you take a rectangular block of glass (n = 1.52) and place a drop of the liquid on the top surface of the block. You shine a laser beam with wavelength 638 nm in vacuum at one side of the block and measure the largest angle of incidence θa for which there is total internal reflection at the interface between the glass and the liquid (Fig. P33.58). Your results are given in the table: What is the refractive index of each liquid at this wavelength?

A ray of light is incident on a plane surface separating two sheets of glass with refractive indexes 1.70 and 1.58. The angle of incidence is 62.0, and the ray originates in the glass with n = 1.70. Compute the angle of refraction.

A ray of light traveling in a block of glass (n = 1.52) is incident on the top surface at an angle of 57.2 with respect to the normal in the glass. If a layer of oil is placed on the top surface of the glass, the ray is totally reflected. What is the maximum possible index of refraction of the oil?

(a) At what angle above the horizontal is the sun if sunlight reflected from the surface of a calm lake is completely polarized? (b) What is the plane of the electric-field vector in the reflected light?

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