5. Exercises

1. Light and Radiation

Electromagnetic Waves (2/4)

Question: Time lapse of sunlight and moonlight reaching the earth

a) The distance between the sun and the earth is approx. 150·10⁶ km. How much time does it take (in seconds and in minutes) for the light emitted by the sun to reach the earth?

b) The distance between the moon and the earth is 384 000 km, hence much shorter. How much time does it take for the sunlight reflected by the moon surface to reach the earth?

Answer:

a) The time lapse of sunlight to reach the earth is

t = \frac{x}{c} = \frac{150 \cdot 10^{9} m}{0.3 \cdot 10^{9} m/s} = 500 s

or 8.33 minutes.

b) For the moonlight it is

t = \frac{x}{c} = \frac{0.384 \cdot 10^{9} m}{0.3 \cdot 10^{9} m/s} = 1.28 s

Question: The light year and the distance to Alpha Centauri

The closest star to the sun is called Alpha Centauri. Its distance is given as 4.247 light years. A light year is the distance travelled by light in one year, a unit of length often used in astronomy. How many kilometres is a light year? How far away in kilometres is Alpha Centauri? You will realise that light years are a very practical unit of length to avoid making distances to distant celestial bodies too unwieldy!

Answer:

A light year corresponds to the product of the speed of light and the time 1 year, i.e. the time 365 days with 24 hours, 60 minutes and 60 seconds each, i.e. 31.536·10⁶ seconds.

0.3 \cdot 10^{9} m / s \cdot 31.536 \cdot 10^{6} s = 9.461 \cdot 10^{15} m

or 9.461·10¹² km or approx. 9.5 trillion kilometres. The distance of 4.247 light years to Alpha Centauri corresponds to approx. 40 trillion kilometres.

Electromagnetic Waves (4/4)

Worksheet: Light and radiation

Please test and deepen your knowledge on the topic of this chapter with the worksheet Light and Radiation. It can also be used as classroom task or homework assignment.

Photons (4/4)

Exercise: A flash of 1000 photons

In the dark of night, we see the flash of a camera far away. We assume that 1000 photons of the flash enter the retina through the pupil of our eye, a really very weak flash of light. In the left-hand column below it says that a dark-adapted eye can probably just barely perceive this flash.

The diameter of the pupil is 4 mm. The light wavelength is 500 nm (a flash is actually white). The flash duration is 1 ms, which corresponds to the flash duration of photo flashes.

a) Calculate the energy of a single photon in electron volts and of the flash consisting of 1000 photons in joules.

b) The photons are evenly distributed over the duration of the flash. What instantaneous power passes through the pupil?

c) What is the irradiance in W/m² of the flash at our location?

Answers:

a) The energy of a photon is:

\begin{array}{l} E = h f = \frac{h c}{λ} \\ ​ ​ ​ = \frac{6.6 \cdot 10^{- 34} Js \cdot 3.0 \cdot 10^{8} m / s}{500 \cdot 10^{-9} m} = 0.4 \cdot 10^{- 18} J \end{array}

1 J corresponds to 1.6·10^-19 eV. It becomes: E = 2.48 eV. Compare the result with the diagram on page Photons (3/4).

The total energy of the 1000 photons of lightning in the eye is 0.4·10^-15 J or 0.4 fJ. (10^-15 = 1 quadrillionth = 1 femto, abbreviation f)

b) The flash energy occurs in one millisecond. The power through the pupil during the flash duration is:

P = \frac{0.4 \cdot 10^{- 15} J}{10^{- 3} s} = 0.4 \cdot 10^{- 12} W

or 0.4 pW. (10^-12 = 1 trillionth = 1 pico, abbreviation p)
It's hard to believe that the eye can see such lightning!

c) The radius of the pupil is r = 2 mm. The power passing through is 0.4·10^-12 W. The irradiance is thus:

\frac{P}{π r^{2}} = 32 \cdot 10^{- 9} \frac{W}{m^{2}}

or 32 nW/m². For comparison: In a cloudless sky, the irradiance on the ground at full moon is about 3 mW/m², the sun at the zenith radiates with about 1 kW/m².

Spectral analysis: Colour glass filters (4/4)

Exercise: Reflectances

Compare the black-and-white images of bands 1 to 4 with the false colour image.

Evaluate the reflectances of vegetated and non-vegetated land surfaces and of the seawater based on their brightness.
Have a look at the table of TM bands on the preceding page. Can you confirm the descriptions unter column characteristics?
Investigate your findings further by comparing with information in related SEOS pages on

spectroscopic characteristics of vegetation in tutorial Remote Sensing and GIS in Agriculture
spectroscopic characteristics of seawater in the tutorial Ocean Colour in the Coastal Zone
classificaton of satellite images in tutorial Classification Algorithms and Methods

No answers are suggested for this task.

Equations ↓

Supplement 1.1: Maxwell's Equations

Question 1: Divergence and curl of a vector

Please show that the equations given above for the divergence $\nabla \cdot \vec{E}$ and the curl $\nabla \times \vec{E}$ of the electric field vector are correct. Are these terms scalar quantities or vectors?

Answers:

\nabla \cdot \vec{E} = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}) \cdot (E_{x}, E_{y}, E_{z}) = \frac{\partial E_{x}}{\partial x} + \frac{\partial E_{y}}{\partial y} + \frac{\partial E_{z}}{\partial z}

is a scalar,

\begin{matrix} \nabla \times \vec{E} = | \begin{matrix} i & j & k \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ E_{x} & E_{y} & E_{z} \end{matrix} | \\ = (\frac{\partial E_{z}}{\partial y} - \frac{\partial E_{y}}{\partial z}, \frac{\partial E_{x}}{\partial z} - \frac{\partial E_{z}}{\partial x}, \frac{\partial E_{y}}{\partial x} - \frac{\partial E_{x}}{\partial y}) \end{matrix}

is a vector. $i$ , $j$ and $k$ are the unit vectors in the direction of the $x$ , $y$ and $z$ coordinates.

Question 2: Gradient of a scalar quantity

To be complete, the ∇ operator can further be applied with scalar quantities. Let $φ (x, y, z)$ be a scalar function in space, e.g. the electric potential. The term ∇φ then denotes the spatial derivative of φ, also called grad φ. ∇φ is a vector since the vector ∇ is multiplied with the scalar φ.

Please write ∇φ in components along the x, y, and z directions.

Answer:

\nabla φ = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}) φ = (\frac{\partial φ}{\partial x}, \frac{\partial φ}{\partial y}, \frac{\partial φ}{\partial z})

Supplement 1.2: The Laplace operator

In Cartesian coordinates, the nabla operator

\nabla = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z})

is the first order spatial derivative. The scalar product $\nabla \cdot \nabla = Δ$ is then the second order spatial derivative, the Laplace operator.

a) Please show that the Laplace operator in Cartesian coordinates is:

Δ = \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}}

b) Please write the wave equation of the electric field in its components in Cartesian coordinates.

Answers:

a) $\nabla \cdot \nabla = Δ = (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial x}) \cdot (\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial x}) = \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} + \frac{\partial^{2}}{\partial z^{2}}$

b) For the electric field $\vec{E} = (E_{x}, E_{y}, E_{z})$ of a wave in any direction is the x component:

\frac{\partial^{2} E_{x}}{\partial x^{2}} + \frac{\partial^{2} E_{x}}{\partial y^{2}} + \frac{\partial^{2} E_{x}}{\partial z^{2}} = ε_{o} μ_{o} \frac{\partial^{2} E_{x}}{\partial t^{2}}

and correspondingly for the y and z components and the magnetic field components.

Question 2: Third Maxwell equation in Cartesian components

Please show by calculus that the field vectors

\vec{E} = (0, E_{y}, E_{z})

und

\vec{B} = (0, B_{y}, B_{z})

applied to the third Maxwell equation yield the components:

\frac{\partial B_{x}}{\partial t} = 0

\frac{\partial B_{y}}{\partial t} = \frac{\partial E_{z}}{\partial x}

\frac{\partial B_{z}}{\partial t} = - \frac{\partial E_{y}}{\partial x}

Answer:

The third Maxwell equation

\nabla \times \vec{E} = - \frac{\partial \vec{B}}{\partial t}

with

\vec{E} = (0, E_{y}, E_{z})

and

\vec{B} = (0, B_{y}, B_{z})

In components:

((\frac{\partial E_{z}}{\partial y} - \frac{\partial E_{y}}{\partial z}), - \frac{\partial E_{z}}{\partial x}, \frac{\partial E_{y}}{\partial x}) = - (0, \frac{\partial B_{y}}{\partial t}, \frac{\partial B_{z}}{\partial t})

The assertion follows from this.

Supplement 1.4:
Energy and Intensity of Electromagnetic Waves

Question 1: Electric and magnetic field of solar radiation

Chapter 2 Thermal radiation shows that the irradiance of solar radiation at the outer edge of the Earth's atmosphere is 1361 W/m², which is known as the solar constant. Please calculate the electric and magnetic field strength of this radiation.

Answer:

⟨ S ⟩ = E_{r a d} = \frac{c ε}{2} E_{o}^{2} = \frac{c}{2 μ} B_{o}^{2} = 1361 \frac{W}{m^{2}}

With $c = 3 \cdot 10^{8} m / s$ , $ε_{o} = 8.854 \cdot 10^{- 12} \frac{A s}{V m}$ and $μ_{o} = 1.26 \cdot 10^{- 6} \frac{V s}{A m}$ follows:

E_{o} = 1013 V / m

B_{o} = 3.38 \cdot 10^{- 6} \frac{N}{C m / s} = 3.38 \cdot 10^{- 6} Tesla

The electric field is very strong, the magnetic field quite small. The cause is - as shown in the supplement - the relationship $E = c B$ . However, the field energies of the electric and magnetic fields are the same.

The amplitude of a plane monochromatic wave is calculated here, which does not apply to solar radiation with its broad spectrum; the numerical values of the result are nevertheless valid.