Snel's Law and Refraction Index of Refraction for Air, Schemes and Mind Maps of Law

Download Snel's Law and Refraction Index of Refraction for Air and more Schemes and Mind Maps Law in PDF only on Docsity! ESCI 340 - Cloud Physics and Precipitation Processes Lesson 13 - Atmospheric Optical Phenomena Dr. DeCaria References: One of the best sources for information about atmospheric optics is the Atmospheric Optics website, http://www.atoptics.co.uk Snel’s Law and Refraction • The index of refraction for a medium is defined as m = c/c̃, (1) where c is the speed of light in a vacuum, and c̃ is the speed of light in the medium.1 • As light passes from one medium into another, there is both reflection and refraction. • Refraction occurs because the wave fronts bend as they cross from one medium into another, causing a ray of light to bend. The ray bends toward the medium that has the slower speed of light (highest index of refraction). • The bending of the ray is quantified by Snel’s Law, which is stated mathematically as sin θ1 sin θ2 = m2 m1 , (2) where θ1 is the angle of incidence (and reflection), θ2 is the angle of refraction, and m1 and m2 are the indices of refraction in the two mediums (see Fig. 1). • The amount by which a ray of light is deflected due to refraction can be quantified in one of two ways. – The bending angle , θ′, is the interior angle between the initial and final rays. – The deviation angle , θ′′, is the complement of the bending angle, θ′′ = 180◦−θ′. – The relationship between bending angle and deviation angle is illustrated in Fig. 2. Index of Refraction for Air • Light travels faster through warm air than it does through cold air. Therefore, the index of refraction for visible light in air decreases with increasing temperature. • Figure 3 shows m− 1 as a function of temperature for yellow light. 1Technically the index of refraction is a complex number, with the real part defined by (1), and the imaginary part linked to the extinction or absorption of light in the medium. 1 Figure 1: Illustration of Snel’s Law. Figure 2: Relationship between bending angle, θ′, and deviation angle, θ′′. • The index of refraction for visible light in air decreases with increasing wavelength. • Figure 4 below shows m− 1 as a function of wavelength. • 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, and will therefore bend more when passing from one medium to another. Mirages • Mirages are caused by the bending of light rays either upward or downward near the ground. • If the air near the ground is very hot, the light rays will bend upward. 2 Figure 5: Path of light through a droplet for a primary rainbow. • There are an infinite number of possible incidence angles, and therefore paths, through the droplet. Figure 7 shows the bending angle as a function of incidence angle, calcu- lated using (5). – Note from Fig. 7 that there is a certain incidence angle, θ̃i, at which the bending angle has a maximum value. – At this particular incidence angle slight changes in θi lead to no change in θ′. – At this special incidence angle, the light beam will be very concentrated. At other incidence angles, the light beam will not be as concentrated. – It is only for this special incidence angle that a primary rainbow will be visible. – θ̃i is found by taking ∂θ′/∂θi = 0 and solving for θi. The result is cos2 θ̃i = 1 3 ( m2 w m2 a − 1 ) . (6) • If we use the value θ̃i in (5), then we obtain the value of the bending angle for our rainbow, which we will call θ̃′. • Since mw is around 1.33 while ma is around 1.0003, we can without much error, use mw ma ∼= mw. Therefore, the equations for the incidence angle and bending angle for the primary rainbow (from (5) and (6)) are often written as cos2 θ̃i = 1 3 (m2 w − 1) (7) 5 Figure 6: Diagram used for calculating (4). and θ̃′ = 4 arcsin ( 1 mw sin θ̃i ) − 2θ̃i. (8) • Table 1 shows the index of refraction for water for four different colors of light, and the associated values for the incidence and bending angles for a primary rainbow. The bending angle of red light is about 42.5◦, while that for violet is about 40.7◦. Violet Green Orange Red λ (µm) 0.4047 0.5016 0.5893 0.7061 mw 1.3427 1.3364 1.3330 1.3300 θ̃i 58.9◦ 59.2◦ 59.4◦ 59.6◦ θ̃′ 40.7◦ 41.6◦ 42.1◦ 42.5◦ Table 1: Index of refraction, incidence angle, and bending angle for various colors of a primary rainbow. • Because red light has a larger bending angle than does violet light, an observer perceives that a primary rainbow has red on the outside, and violet on the inside (see Fig. 8). – Because each color bends through a slightly different angle, each color range you observe from a rainbow is from a different set or population of droplets. The violets you observe could not have been refracted through the same droplets as the reds you observe. 6 Figure 7: Primary rainbow bending angle as a function of incidence angle. • If the Sun is more than 42.5◦ above the horizon then a primary rainbow cannot be seen by a ground-based observer. Secondary Rainbows • It is also possible to have a path through the water droplet that has two internal reflections, such as that shown in Fig. 9. • The relation between the bending angle, θ′, and the incidence angle, θi, is worked out by deducing the following four independent relationships from Fig. 9, θi/2 + α + β = 180◦ (9) θt + γ = 90◦ (10) 3γ + β = 180◦ (11) θi + α = 180◦, (12) and eliminating the angles α, β, and γ from (9)–(12) to get θ′ = 180◦ + 2θi − 6θt. (13) – From Snel’s Law we have sin θi sin θt = mw ma ∼= mw, (14) or θt = arcsin ( 1 mw sin θi ) . (15) 7 Figure 10: Bending versus incidence angle for primary and secondary rainbows. There are no bending angles between 42◦ and 50◦ for either type of rainbow, resulting in relative darkness between the bows. – Higher order bows are much fainter, because every internal reflection results in a loss of light from the beam. – Many higher order bows also have bending angles of more than 90◦, meaning you have to be looking toward the sun to see them, which makes them more difficult to see. • Supernumerary rainbows are concentric arcs that appear on the inside of a primary rainbow. – Supernumeraries appear very close to each other, and are often greenish or pur- plish. – If two rays of the same color take very slightly different paths through the droplet, then they may either constructively interfere with each other, enhancing the brightness, or they may destructively interfere with each other, decreasing the brightness of the beam. – To form supernumeraries the droplets must be fairly uniform, nearly spherical 10 raindrops (so therefore, small droplets). • Sometimes the top of a rainbow will appear to be ’twinned’. This is thought to be due to the nonspherical shape of large raindrops as they fall, causing the light rays to bend through a different angle than they would through a spherical drop. • See http://www.atoptics.co.uk for more detailed information on higher order, supernu- merary, and other types of rainbows. Corona • Corona appear as colored rings around the sun or moon. • They are smaller than the 22◦ halo. • Corona are formed by diffraction of light through cloud droplets, and sometimes through cloud ice crystals. Ice Crystals as Prisms • Ice crystals come in many different shapes, or habits (plates, columns, needles, etc.). – The habit depends on the temperature and humidity at which the crystal forms. – Most ice crystals are six-sided (hexagonal). • Ice crystals act as prisms, refracting the light as it passes through the crystal. • Figure 11 shows some possible symmetric paths of light through a hexagonal column. Figure 11: Paths of light through a hexagonal ice crystal column or plate. • One path is for the light to pass laterally across the crystal, shown on the left of Fig. 11. – In this case, the crystal acts like a prism with an angle of 60◦. • The other possible path is for the light to pass longitudinally along the crystal, as shown on the right of Fig. 11. 11 – In this case, the crystal acts like a prism with an angle of 90◦. • A third path, shown in Fig. 12, is impossible for ice crystals because the index of refraction for ice is too large. Figure 12: Invalid path of light through a hexagonal ice crystal column. • The deviation angle θ′′ from a prism depends on the incidence angle θi, the index of refraction of the prism, m, and the angle of the prism, A. The formula is θ′′ = θi − A+ arcsin ( m sin [ A− arcsin ( sin θi m )]) , (19) the derivation of which is shown in the Appendix. • Figure 13 shows the relationship between deviation angle and incidence angle from (19) for the two possible paths through an ice crystal for three different wavelengths of light; 0.70µm, 0.58µm, and 0.40µm. The indices of refraction for these wavelengths are 1.3070, 1.3100, and 1.3195.2 • Note that there is a minimum deviation angle at which ∂θ′′/∂θi = 0, and for which the beam through the prism will be concentrated. This concentrated path also hap- pens to be the path that passes symmetrically through the prism (also derived in the Appendix). • The minimum deviation angle is given by θ′′min = 2 arcsin [ m sin ( A 2 )] − A. (20) • The index of refraction for ice is wavelength-dependent, and is larger for shorter wave- lengths. – Because of this wavelength dependence, red light has the smallest deviation angle, while violet light has the largest deviation angle. – When sunlight refracts through ice crystals, the red light appears on the side of the phenomenon that is closer to the sun, while the blue/violet light appears on the side further from the sun. 2Data are from Warren, S. G., 1984: Optical constants of ice from theultraviolet to the microwave. Appl. Opt., 23, 1206-1224. 12 • Finally, from Snel’s law we have α = arcsin ( sin θi m ) . (26) so that (25) becomes θ′′ = arcsin ( m sin [ A− arcsin ( sin θi m )]) − A+ θi. (27) • A plot of deviation angle versus incidence angle, Fig. 13, shows that there is only one particular incidence angle for which the deviation is minimized. To find this incidence angle we could differentiate (27) with respect to θi, set the result equal to zero, and then solve for θi; however, this would be very tedious. • Instead, we can recognize that at the minimum deviation, θi must equal θo. For if it weren’t, then if we reversed the ray in Fig. 17 the angle of incidence would be θo, and yet we would get the same deviation as for the forward ray. So, there would be two incidence angles giving us the same deviation! That does not square with our graph, and so we can deduce that at minimum deviation, θi = θo. Therefore, it is the symmetric path through the prism that gives the minimum deviation of the ray. • For this symmetric path we have α = β = A/2, and so from (22) we get θ′′min = 2θi − A. (28) • From Snel’s Law, sin θi = m sinα = m sin ( A 2 ) , or θi = arcsin [ m sin ( A 2 )] . (29) • Putting (29) into (28) gives θ′′min = 2 arcsin [ m sin ( A 2 )] − A. 15 Figure 13: Deviation angle versus incidence angle for ice crystals. The three colored lines are for wavelengths corresponding to red, yellow, and violet light (0.70µm, 0.58µm, and 0.40µm). The minimum deviation angle for yellow light (shown by the dashed horizontal lines) is either 22◦ or 46◦, depending on which path through the crystal the ray traverses. 16 Figure 14: Schematic showing various optical phenomena associated with ice crystals: a. parhelia (sun dogs); b. 22◦ halo; c. 46◦ halo; d. sun pillars; e. upper-tangent arc; f. cir- cumzenithal arc; g. lower-tangent arc; h. parhelic circle; i. infraleral arc; j. supralateral arc. Figure 15: Path of light through a horizontally-oriented hexagonal plate, resulting in forma- tion of a circumzenthal arc. 17