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## Physical Meteorology

by: Cordie Miller

62

0

16

# Physical Meteorology ESCI 340

Cordie Miller

GPA 3.65

Alex DeCaria

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COURSE
PROF.
Alex DeCaria
TYPE
Class Notes
PAGES
16
WORDS
KARMA
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## Popular in Earth Sciences

This 16 page Class Notes was uploaded by Cordie Miller on Thursday October 15, 2015. The Class Notes belongs to ESCI 340 at Millersville University of Pennsylvania taught by Alex DeCaria in Fall. Since its upload, it has received 62 views. For similar materials see /class/223514/esci-340-millersville-university-of-pennsylvania in Earth Sciences at Millersville University of Pennsylvania.

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Date Created: 10/15/15
ESCI 340 Physical Meteorology Cloud Physics Lesson 7 Atmospheric Optics Dr DeCaria References An Introduction to Atmospheric Radiation Liou Atmospheric Science An Introductory Survey Wallace and Hobbs Reading Petty Chapter 4 Skip 413 and 414 INDEX OF REFRACTION The index of refraction sometimes called the refractive index is a complex number m m imi 0 The real part m is what we usually work with and is the ratio of the speed of light in a vacuum cm to the speed of light in the medium in question c my cm c 0 The imaginary part m is related to the absorption of radiation by the medium and can be related to the absorption or massabsorption cross sections via kl Oan47rmi AP A A When light passes from one medium to another the index of refraction changes and hence so does the speed of light A ray is defined as a line perpendicular to the wave front As the light passes from one medium into another there is both re ection and refraction 0 The wave fronts bend as they cross from one medium into another Therefore a ray will bend as it crosses from one medium into another 0 The amount a ray is bent is given by Snel s Law which states that 51nd1 m2 sin 02 m1 where 01 is the angle of incidence and re ection and 02 is the angle of refraction see diagram below maiden my re uxed my refraczed my INDEX OF REFRACTION FOR AIR 0 The index of refraction for Visible light in air depends on both the wavelength of the light and on the temperature of the air The real part of the index of refraction for Visible light in air decreases with increasing temperature 0 The diagram below shows m 1 as a function of temperature for yellow light Index of refraction for yellow light A 3010 F 101325 mb 2810quot 7 7 a 2510quot 7 7 Er 2410quot 7 7 2210quot 7 2010quot 0 100 0 40 60 80 Temperature C 0 Though the change in index of refraction with temperature is slight it is enough to be important Therefore 0 Light travels faster in warmer air and slower in colder air 0 Because the index of refraction is larger in colder air from Snell s law we deduce that light bends toward colder temperatures The real part of the index of refraction for Visible light in air decreases with increasing wavelength 0 The diagram below shows m 1 as a function of wavelength Tndex of refraction for Visible light 27810 1 101325 mb 27610 T 20 C 39 27410quot 7 7 Er 27210quot 7 7 27010quot 04 o 5 o 6 0 7 wavelength microns 0 The change of the index of refraction in air over the visible wavelengths is very small but shorter wavelengths do travel more slowly than do longer wavelengths SCATTERING The extinction cross section of an object consists of a part due to absorption and a part due to scattering 71 a a The scattering cross section can be written as 02 KA where A is the geometric physical crosssection of the object and K is the scattering area coef cient or scattering ef cien cy 0 The scattering e ciency is the ratio of the scattering crosssection to the geometric gross section of the particle The scattering ef ciency will depend on the size of the particle the wavelength of light and on the index of refraction of the particle We can combine the wavelength and particle size into one parameter called the size parameter de ned as as 27ml because it is the ratio of the particle size to the wavelength of light that is important GEOMETRIC OPTICS REGIME 0 If the size parameter is greater than 50 or so the K approaches an asymptotic value of 2 and the scattering can be described using geometric optics In this regime all wavelengths are scattered equally so the scattered light will be of the same mixture of wavelengths as the incident light This is why clouds appear white 0 RAYLEIGH SCATTERING REGIME O 0 If the size parameter is small altlt 1 then the scattering efficiency is given by 2 K m 14 6 mr1 For a scatterer of a given size this means that K olt 1394 and that shorter wavelengths are scattered much more efficiently than are longer wavelengths For visible light the ratio of KblueKred is about 35 so that blue light is scattered nearly 4 times more efficiently than is red light This is why the sky appears blue 0 MIE SCATTERING REGIME O O O In between the two extremes of the Rayleigh and geometric optics regimes scattering is under the more complex Mie regime In this regime the scattering efficiency exhibits oscillatory behavior with increasing size parameter In the Mie regime if the particles are nearly uniform in size the scattered light may either appear blue or reddish depending on whether K is increasing or decreasing with a If the particles are not uniforms ie there is a whole spectrum of sizes present then several minima and maxima will be included and the light will appear whitish MIRA GES O Mirages are caused by light bending due to changes in refractive index due to temperature changes along the path of the ray 4 0 Because light bends toward colder temperatures if the air near the ground is very hot the light rays will bend upward 0 This causes objects far away to appear inverted and underneath their actual position 0 This results in an inferior image 0 This is what causes hot asphalt to appear wet up ahead because the sky appears as an inferior image in the road If the air near the ground is very cold the light rays bend downward 0 This causes far away objects to appear above their actual position 0 This results in a superior image 0 This can actually allow objects that are over the horizon to be seen PRIMARY RAINBOWS drops Rainbows are caused by the refraction and re ection of light by rain and cloud 0 The index of refraction for water is much larger than that for air and also depends on wavelength 0 The index of refraction is greater for shorter wavelengths so therefore shorter wavelengths bend more The figures below shows the path of a light ray through a spherical water droplet for a primary rainbow a Sun 395 my Incident Sun 395 my out The angle 9 is the incidence angle I The interior angle between the incoming and outgoing ray 9 j is called the bending angle I The exterior angle 9 is called the deviation angle 0 Using Snel s Law and geometry the angle 9 for the primary rainbow can be related to the incidence angle 9 as follows refer to diagram below 5 Sun 395 my Incident Sun 395 my out 0 From the diagram and from Snel s law we have the following relations 9 9 a 9 2 2 180quot 9 180quot 9 amsini sin 9 m where we de ne the index of refraction ofwater with respect to air m as m E mvwmevmvalv 0 Solving this system of equations for 9 gives 9 4amsin isin 917 29 Bending anglefor primary rainbow 1 m mL 4 w number 39 and thereforepaths through the droplet The plot below shows the deviation and bending angles as a function ofincidence angle Primary Rainbow 200 7 t t 7 deviation angle 1507 7 100 i 7 degrees bending angle 0 V l l l l 0 20 40 60 80 incident angle degrees 100 0 Note that there is a certain incidence angle at which the deviation is a minimum and therefore the bending angle is a maximum At this incidence angle slight changes in the incidence angle 0 lead to no change in QC 0 At this special incidence angle the light beam will be very concentrated At other incidence angles the light beam will not be as concentrated 0 It is only at the this special incidence angle where the deviation angle is a minimum and the bending angle is a maximum that a rainbow will form To find this special incidence angle analytically we need to differentiate equation 1 with respect to 0 set it equal to zero and solve for 0 this is left as an exercise Doing so shows that for the primary rainbow cos2 m2 13 Incidence angle for primary rainbow 2 0 Equations 1 and 2 are used together to determine the angle that a primary rainbow subtends The value of m depends on the wavelength of the light The table below shows these values for a few colors violet 04047 Color Man m 13427 red 07061 13300 green 05016 orange 05893 13364 13330 Since different colors have different indices of refraction then each color will have a slightly different angle after it leaves the droplet Thus the droplets act as prisms to separate the colors O For red light the angle is approximately 42 This is the angle that a primary rainbow subtends 0 The shorter wavelengths have a smaller 9 because they have a larger index of refraction Therefore red appears as the outer band of a primary rainbow with the other colors following in order of decreasing wavelength 0 If the Sun angle is larger than 42 a primary rainbow cannot be seen by a groundbased observer SECONDARY RAINBOWS 0 It is also possible to have a path through the water droplet that has two internal reflections such as that shown below Sun 395 my out 0 For this case 19 we have the following relations a a 180 e 2 or 180 37 180 7 19 90 19r arcsin i sin 8 m 0 Solving this system of equations for 9 gives 19 180 219 7 Garcsini sin 19 j Bending angle for secondary rainbow 3 m 0 As before there are an infinite number of such paths However there is one special path for which d fd l 0 which gives an incidence angle of cos2 8 m2 7 18 Incidence anglefor secondary rainbow 4 O For red light the angle 9 is approximately 50 This is the angle that a secondary rainbow subtends O The shorter wavelengths have a larger 0 in this case 0 Therefore red appears as the inner band of a secondary rainbow with the other colors following in order of decreasing wavelength 0 Secondary rainbows are much dimmer than primary rainbows because every time the light refracts andor re ects some of the radiance is lost out of the beam 0 If the Sun angle is larger than 50 a secondary rainbow cannot be seen by a ground based observer PRISMS O A prism is a transparent wedge of material which allows light to pass through it and which separates the colors of the light 0 There are many paths that the light can take through a prism One such path is shown below 0 For a prism 0 is the deviation angle 0 We can find the angle by which the light ray is bent through a prism by using the following relationships from the diagram aQ Q r 62 y180 a 180 y from which we can deduce that yes 9 9192 From the diagram we can also see that 180 A180 1 2 1 2A 5 so that 39 A 6 The gure below shows a plot of the deviation angle as a function of angle of incidence for a 70 prism with a refractive indeX of 12 40 w o H H m N o WW WHWHW deviation angle degrees o m Hm 0 l l l l 0 20 40 60 80 incident angle degrees 100 0 As before there are an infinite number of paths through the prism but there is only one for which the beam will be concentrated ie for which an infinitesimal change in 0i will produce no change in 0 This will occur when the deviation angle is a minimum just like for the rainbow 0 We find this angle by taking d 639 d 6 z 0 Using equation 6 we get d7 1 d6 01 1 7 Mi gt 0 From the chain rule and equation 7 we have d0 2 day 102 24 8 101 102 101 101 0 From Snel s law we know sin 01 sin 0 m m sin 01 sin 02 which when differentiated gives d0i zmcosg1 d0 zmcosg2 9 d 01 cos 0i 102 cos 0 Also since 01 02 A equation 5 we have 1 10 101 Using 9 and 10 in 8 yields cos 02 cos 0i 11 cos 01 cos 0 39 O The only way for condition 11 to be true is for the following to also be true cos 0i cos 0 cos 01 cos 0239 This leads to the important conclusion that for a beam of light to pass through a prism without significant divergence of the light rays the angle of incidence must equal the angle at which the beam leaves the prism 0 Another way of saying this is that at the 39 39 devia nn angle the beam passes symmetrically through the prism as shown below 0 The deviation angle in this symmetric case from 6 is H 29 7A 12 0 The angle of incidence is going to depend on the angle of the prism A and the index of refraction m 0 We know from equation 5 that H A 2 and from Snel s Law that sin 9 m sln 9 which combined show that sin 9 msinA2 13 and therefore 039 ZaIcsinm sinA27 A Minimum deviation angle through aprism 14 REFRACTION OF LIGHT THROUGH ICE CRYSTALS 0 Ice crystals come in many different shapes or habits 0 The habit depends on the temperature at which the crystal formed 0 However all ice crystals are sixsided hexagonal 0 Optical phenomenon like halos and sun dogs are associated with columnar or platelike ice crystals like that shown below 12 so 0 There are two possible symmetric paths of the light through the column and these are shown below v60 0 One path is for the light to pass laterally across the cyrstal such as shown on the left I In this case the crystal acts like a prism with an angle of 60 and using the index of refraction for ice m 131 equation 14 gives an angle of bending of about 22 0 The other possible path is for the light to pass longitudinally along the crystal such as shown on the right I In this case the crystal acts like a prism with an angle of 90 and using the index of refraction for ice m 131 equation 14 gives an angle of bending of about46 0 A third path shown below is impossible for ice crystals because the index of refraction for ice is too large and therefore equation 13 gives a nonsensical result for sin 9 HALOS SUN DOGS AND SUN PILLARS Halos sun dogs also called perihelia and sun pillars are caused by the re ection or refraction of sunlight from ice crystals Halos and sun dogs are displaced either 22 or 46 from the Sun depending on which path the light takes through the ice crystal see below v60 0 Halos are arcs around the Sun and occur when the ice crystals are randomly oriented 0 Sun dogs only appear to either or both sides of the sun and occur when the ice crystals are primarily oriented vertically 0 Sometimes halos and sun dogs appear together 0 Halos of other dimensions are sometimes observed as well as tangent and circumzenithal arcs and are associated with oddly shaped ice crystals which have a pyramidal cap or to spinning ice crystal Sun pillars are caused by re ection rather than refraction and appear as a column of light above and sometimes below the sun o For sun pillars the ice crystals are more platelike rather than columnar and are oriented with the at platelike surfaces horizontally They then act together like a large rnirror crystals 0 a are run dog 0 h is the 22 halo 0 c is the 46 Bhan o d are run pillar o e is the upper tangent m o f is the eiyeumzenitlml m o g is the Lower tangent m o h is the perhelt39e et39yele EXERCISES 1 Squot 5 5quot Take d d 0 of equations 1 and 3 and set it equal to zero to derive equations 2 and 4 respectively You are on a planet where the rain drops are made up of a liquid with an index of refraction of 144 You notice a rainbow Assuming it is a primary rainbow what is the bending angle If ice crystals were pentagonal 5sided columns list the angles of all possible haloes You are on a planet with strange clouds You suspect the clouds are made up of some sort of crystal and other measurements lead you to believe that the index of refraction of the crystal is 105 You have observed halos of 24 59 and 169 What would you guess the shape of the crystals are You are on a planet where the clouds are made up of droplets of an unknown liquid You notice a rainbow assume it is a primary rainbow with an angle of 150 What is the index of refraction to two decimal places of the cloud droplets

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