Solutions For Homework #1 - Stanford University

1 Solutions For Homework #1 Problem 1:Ifcis the speed of light, is the wavelength andfis the frequency, theequationc= fcan be used to calculate one of the the three values, if thetwo others are given. The speed of light is defined to be exactly299792458 m/s( 3 108m/s). Thus, for a frequency off= 10 GHz = 10 109Hz, =cf=299792458 m/s10 GHz 3 108m/s1010Hz= = 3 cmand for a wavelength of = 25cm= =299792458 12 108Hz = GHzProblem 2:The boiling temperature of water at sea level is100oC. To convert this todegrees Kelvin, just add :100oC= ( + 100)K= 3:The formula for the wavelength of the energy emission peak is max=2898 mKTwhereTis the black-body temperature. Using the temperatures for the surface ofthe sun (T= 6000K) and an effective value for the earth (T= 255K), we m1 Because the reflected radiation has the same spectrum as the light incidentfrom the sun, the peak wavelength is also the same, 4:Using the equation from problem 3 and the surface temperature of venus(TV enus= 750K), we get max=2898 mK750K= m05101520250500100015002000250030003500w avelength ( m) emitted spectral radiance (W/m2/ )Blackbody curve for Venus at T = 750 Kpeak poweroccursat wavelength = mFigure 1: Blackbody curve of Venus at a temperature ofT= emmitted spectral radiance values were calculated using Planck s for-mula:S( ) =2 hc2 51ech/ kT 1 Problem 5:The industrial pollutant would not be a concern for global warming.

2 The prin-ciple atmospheric windows in the thermal infrared band are between3 5 mand28 14 m. Wavelengths in these windows absorb the most energy which is trans-mitted through the Earth s atmosphere, and therefore contribute the most to thegreenhouse effect. The pollutant in question absorbs radiation at a28 mwave-length, which is not in or near the atmospheric window and would not contributeto the greenhouse :The fact that the28 mwavelength does not coincide with either thesun s or the earth s peak emission wavelength does not mean that there is no sig-nificant amount of radiation at this 6:Estimate how much the sea level would rise if the entire Antarctic ice cap wereto melt by converting the volume of the ice cap to its water equivilant and thendividing the volume by the surface area of the world ocean. The area of Antartica(from encarta website) is approximately 14,000,000km2. Using the approximateice cap thickness of 3000 m (3 km):area= (14,000,000km2)(3km) = 42,000,000km3 Convert the ice volume to water volume:water volume= (42,000,000km3)( ) = 37,800,000km3 The surface area of the world ocean (from Wikipedia website)is approximately361,000, level rise=37,800,000km3361,000,000km2 or105mThis 105 m rise in sea level would put Stanford under water because its elevationat Mitchell is about 32 meters (check out for elevations).

3 Problem 7:The distance in pixels between Hoover tower and Memorial church can becalculated with the Pythagorean theorem:d= (400 300)2+ (250 300)2 112pixels3 The distance in meters (estimated from a campus map) is approximately325me-ters. Thus, the pixel spacing is d=325m112= 8:Your hometown lab exercise4