1 / 27

Lecture 11: Light

Lecture 11: Light. The Cosmic Messenger. Basic Properties of Light. light is a form of energy – radiative energy rate of energy output (from Sun, lightbulb, etc.) is measured in Watts (Joule/s). The speed of light.

Télécharger la présentation

Lecture 11: Light

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Lecture 11: Light The Cosmic Messenger

  2. Basic Properties of Light • light is a form of energy – radiative energy • rate of energy output (from Sun, lightbulb, etc.) is measured in Watts (Joule/s)

  3. The speed of light • Galileo tried to measure the speed of light using people with lanterns – failed to detect any finite speed • Romer discovered evidence that light has a finite speed in 1676 based on observations of eclipses of the moons of Jupiter • measured accurately on Earth by Fizeau and Foucault in 1850: c = 3 x 108 m/s

  4. The Fizeau-Foucault experiment

  5. Frequency and Wavelength • all light always travels at a fixed speed (in a vacuum): c = 3 x 108 m/s • frequency (n) and wavelength (l) are related by the formula: l n = c • units of frequency are cycles per second = Hertz = Hz = (1/s)

  6. Frequency and Wavelength: example problem Find the frequency of the first Balmer line of Hydrogen, which has a wavelength of 656.3 nm. remember 1 nm = 10-9 m, so l = 6.563 x 10-7 m • l n = c  n = c/l • = (3 x 108 m/s) / (6.563 x 10-7 m) • = 4.57 x 1014 Hz

  7. Frequency and Energy What is the energy of the photons that produce this Balmer line? Remember the frequency n =4.57 x 1014 Hz E = h n E = (6.626 x 10-34 J s)(4.57 x 1014 Hz) E = 3.02 x 10-19 J

  8. Properties of Light • light can be characterized by two numbers • frequency or wavelength • intensity or brightness (amplitude)

  9. What is Light? • Most waves propagate through some sort of medium (e.g., water waves, sound waves in air, etc.) • if light is a wave, what does it wave in? • the idea of the ‘luminiferous aether’ was invented to answer this question.

  10. The Aether theory and the Michelson-Morley experiment

  11. Electromagnetic radiation

  12. The electromagnetic spectrum

  13. Matter and Light • light interacts with matter in four ways: • emission • absorption • transmission • reflection/scattering

  14. Matter and Light • materials that transmit light are called transparent • materials that absorb light are called opaque • the degree to which a material absorbs light is called its opacity (high opacity  absorbs more light) • objects appear to have different colors because of the way that they transmit or reflect light

  15. Scattering: why the sky is blue

  16. and sunsets are red

  17. Flux = Energy/Area Asphere = 4pR2

  18. Blackbody Radiation radiation from an opaque body follows two laws: • the energy flux is proportional to the temperature of the object to the fourth power (Stephan-Boltzmann law) F = [5.7 x 10-8 W/(m2 K4)] x T4 • the average energy of the light (photons) emitted is higher for higher temperature objects (so the wavelength is shorter; Wien’s Law) lpeak = (2.9 x 10 6 / T [K]) nm

  19. The Sun as a Blackbody • The peak wavelength of the Sun’s light is about 500 nm. What is the surface temperature of the Sun? • we can use Wien’s law: T = (2.9 x 106 nm)/lpeak = (2.9 x 106 )/(500 nm) T = 5800 K

  20. The luminosity of the Sun is 3.90 x 1026 W. Find the temperature of the Sun. • this time we’re going to use the Stephan-Boltzman law: F = [5.7 x 10-8 W/(m2 x K4)] T4 first we need to find the flux at the Sun’s surface. remember flux = energy/area so Fsun = Lsun/(4pR2sun) Rsun = 6.96 x 108 m  F = 6.41 x 107 W m-2 • now we use T = (F/ 5.7 x 10-8 W/(m2 x K4))1/4  T = 5800 K

  21. Other Stars and our Sun • Sirius is the brightest star in the night sky. It appears blue and its peak flux is at 280 nm, in the UV. • is Sirius hotter or cooler than our Sun? What is its temperature? • compare the energy flux at the surface of Sirius with that at the surface of our Sun.

  22. The End

More Related