Snell's law

12 Key excerpts on "Snell's law"

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  eBook - ePub

  Optics for Materials Scientists

  • Myeongkyu Lee(Author)
  • 2019(Publication Date)
  • Apple Academic Press

    (Publisher)

  This law represents the relationship between the angles of incidence and refraction, when a light wave passes through a boundary between two isotropic media with different refractive indices. Snell’s law can be derived from Fermat’s principle or the application of boundary conditions for electromagnetic waves. The detailed behaviors of reflection and refraction in a given situation are well explained by the Fresnel equations, which describe what fraction of the incident light is reflected and what fraction is refracted (i.e., transmitted) at a planar interface separating two optical media. They also describe the phase shift of the reflected light. It will be shown that all these quantities depend not only on the change in refractive index and the angle of incidence, but also the polarization state of the incident light. This chapter treats the general features associated with the propagation, reflection, and refraction of light in isotropic media. These media are assumed to be linear, homogeneous, and nonmagnetic. As discussed in Chapter 1, the refractive index of a substance varies with the wavelength of incident electromagnetic radiation. All materials have a refractive index very close to 1 at X-ray wavelengths. Therefore, no refraction occurs in this X-ray range. Here, we are concerned with the visible range, in which most transparent materials exhibit a nearly constant refractive index higher than unity. 2.2 LAWS OF REFLECTION AND REFRACTION In geometric optics, the concept of a light ray is useful and also necessary. A ray is a line drawn along the direction of radiant energy flow. Thus, the ray can be regarded as the path along which light energy is transmitted from one point to another. For a plane wave traveling within homogeneous isotropic media, rays will be straight, parallel lines normal to its wavefronts

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  Gemmology

  • P. G. Read(Author)
  • 2013(Publication Date)
  • Butterworth-Heinemann

    (Publisher)

  The refraction of a light ray is slightly more complicated, at least mathematically, and for many years it was not fully understood. Then in 1621 Snell discovered the underlying relationship between incident rays and refracted rays and laid the foundation for the subsequent rapid advances in applied optics. He expressed this relationship in his two laws of refraction: 1. When a light ray passes from one medium into another there exists a definite ratio between the sines of the angle of incidence and the angle of refraction. This ratio is dependent only on the two media and the wavelength of the light. 2. The incident ray, the refracted ray and the normal (at the point of incidence) are all in the same plane. 85 86 Reflection and refraction NORMAL N Figure 9.1 Snell's first law of reflection states that the angle of incidence equals the angle of reflection (ION = NOR) Figure 9.2 (a) The incident light entering a gemstone (at an angle other than 90° to the surface) is refracted towards the normal. Light leaving a gemstone (other than at 90°) will be refracted away from the normal, (b) If air is the less dense medium, the RI of the denser medium is the ratio of the sines of the angles ION and MOR The word refraction simply means angular deflection. When a ray of light passes from one medium (such as air) into an optically denser medium (such as a gemstone), at an angle other than 90°, the ray is refracted or bent towards the normal (see Figure 9.2(a)). Conversely, when the ray leaves the gemstone and passes into the air, it is refracted away from the normal. The greater the difference between the optical densities of the two mediums (or, in the case of a gemstone surrounded by air, the greater the optical density of the gem), the greater will be the amount of refraction.

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  College Physics

  • Michael Tammaro(Author)
  • 2019(Publication Date)
  • Wiley

    (Publisher)

  The dashed line is the normal to the boundary, and θ and θ ′ are the angle of incidence and the angle of refraction, respectively. n n′ h θ′ θ′ θ θ λ λ′ The key to deriving Snell’s law is identifying the two right triangles shaded green and gray. These triangles have the same hypotenuse of length h. The lengths of the sides opposite the angles θ and θ ′ are the wavelengths λ and λ′, respectively, as shown. We can take the sine of each angle as follows: h sinθ λ = and h sinθ λ ′ = ′ 716 | Chapter 26 From Equation 26.1.2, the wavelengths are related to the indices of refraction by / ( ) n n λ λ ′ = ′ . Making this substitution into the second equation, we have / n n h sinθ λ ( ) ′ = ′ or / n n h sinθ λ ′ ′ = Combining this result and the previous result gives / n n sin sin θ θ = ′ ′ Finally, rearranging this equation yields Snell’s law: n n sin sin θ θ = ′ ′ 26.2 Solve problems dealing with total internal reflection. We have seen that a light ray will refract away from the normal when it enters a medium of lower refractive index. This is illustrated in Animated Figure 26.2.1, where a laser is under water and aimed up toward the surface. When the laser is directed vertically upward, the angle of incidence is zero, in which case the angle of refraction is zero, too. As the angle of incidence θ is increased, the angle of refraction θ ′ also increases. There is always partial reflection at the boundary between two transparent mediums, and the reflected ray makes an angle r θ θ = with the normal. 26.2 TOTAL INTERNAL REFLECTION Learning Objective Animated Figure 26.2.1 When a light ray travels from water into air, the ray refracts away from the normal. At the critical angle c θ , the angle of refraction is 90°. For incidence angles greater than the critical angle, all of the light is reflected back into the water. I N T E R A C T I V E F E A T U R E The critical angle c θ is the angle of incidence for which the angle of refraction is ° 90 .

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  University Physics Volume 3

  • William Moebs, Samuel J. Ling, Jeff Sanny(Authors)
  • 2016(Publication Date)
  • Openstax

    (Publisher)

  Since the speed of light is smaller in the second medium, the waves do not travel as far in a given time, and the new wave front changes direction as shown. This explains why a ray changes direction to become closer to the perpendicular when light slows down. Snell’s law can be derived from the geometry in Figure 1.28 (Example 1.6). Chapter 1 | The Nature of Light 29 Figure 1.28 Huygens’s principle applied to a plane wave front traveling from one medium to another, where its speed is less. The ray bends toward the perpendicular, since the wavelets have a lower speed in the second medium. Example 1.6 Deriving the Law of Refraction By examining the geometry of the wave fronts, derive the law of refraction. Strategy Consider Figure 1.29, which expands upon Figure 1.28. It shows the incident wave front just reaching the surface at point A, while point B is still well within medium 1. In the time Δt it takes for a wavelet from B to reach B′ on the surface at speed v 1 = c/n 1 , a wavelet from A travels into medium 2 a distance of AA′ = v 2 Δt, where v 2 = c/n 2 . Note that in this example, v 2 is slower than v 1 because n 1 < n 2 . Figure 1.29 Geometry of the law of refraction from medium 1 to medium 2. 30 Chapter 1 | The Nature of Light This OpenStax book is available for free at http://cnx.org/content/col12067/1.4 1.5 Solution The segment on the surface AB′ is shared by both the triangle ABB′ inside medium 1 and the triangle AA′B′ inside medium 2. Note that from the geometry, the angle ∠BAB′ is equal to the angle of incidence, θ 1 . Similarly, ∠AB′ A′ is θ 2 . The length of AB′ is given in two ways as AB′ = BB′ sin θ 1 = AA′ sin θ 2 . Inverting the equation and substituting AA′ = cΔt/n 2 from above and similarly BB′ = cΔt/n 1 , we obtain sin θ 1 cΔt/n 1 = sin θ 2 cΔt/n 2 . Cancellation of cΔt allows us to simplify this equation into the familiar form n 1 sin θ 1 = n 2 sin θ 2 .

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  Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  If the index of refraction of the surrounding fluid equals that of the glass prism, then n 1 5 n 2 , and Snell’s law (n 1 sin u 1 5 n 2 sin u 2 ) reduces to sin u 1 5 sin u 2 . Therefore, the angle of refraction equals the angle of incidence, and no bending of the light occurs. Related Homework: Check Your Understanding 16 26.6 | Lenses 735 After reflection from the back surface of the droplet, the different colors are again refracted as they reenter the air. Although each droplet disperses the light into its full spectrum of colors, the observer in Figure 26.21a sees only one color of light coming from any given droplet, since only one color travels in the right direction to reach the observer’s eyes. How- ever, all colors are visible in a rainbow (see Figure 26.21b) because each color originates from different droplets at different angles of elevation. Check Your Understanding (The answers are given at the end of the book.) 13. Two blocks, made from the same trans- parent material, are immersed in different liquids. A ray of light strikes each block at the same angle of incidence. From the draw- ing, determine which liquid, A or B, has the greater index of refraction. 14. A beam of violet-colored light is propagating in crown glass. When the light reaches the boundary between the glass and the surrounding air, the beam is totally reflected back into the glass. What happens if the light is red and has the same angle of incidence u 1 at the glass–air interface as does the violet-colored light? (a) Depending on the value for u 1 , red light may not be totally reflected, and some of it may be refracted into the air. (b) No matter what the value for u 1 , the red light behaves exactly the same as the violet-colored light. (Hint: Refer to Table 26.2 and review Section 26.3.) 26.6 | Lenses The lenses used in optical instruments, such as eyeglasses, cameras, and telescopes, are made from transparent materials that refract light.

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  Introduction to Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2015(Publication Date)
  • Wiley

    (Publisher)

  How can the situations illustrated in Figure 26.19 arise? Reasoning and Solution Snell’s law of refraction includes the refractive indices of both ma- terials on either side of an interface. With this in mind, we note that the ray bends upward, or away from the normal, as it enters the prism in Figure 26.19a. A ray bends away from the nor- mal when it travels from a medium with a larger refractive index into a medium with a smaller refractive index. When the ray leaves the prism, it again bends upward, which is toward the normal at the point of exit. A ray bends toward the normal when traveling from a smaller toward a larger refractive index. Thus, the situation in Figure 26.19a could arise if the prism were immersed in a fluid, such as carbon disulfide, that has a larger refractive index than does glass (see Table 26.1). We have seen in Figures 26.18a and 26.19a that a glass prism can bend a ray of light either downward or upward, depending on whether the surrounding fluid has a smaller or larger index of refraction than the glass. It is logical to conclude, then, that a prism will not bend a ray at all, neither up nor down, if the surrounding fluid has the same index of refraction as the glass—a condition known as index matching. This is exactly what is happening in Figure 26.19b, where the ray proceeds straight through the prism as if the prism were not even there. If the index of refraction of the surrounding fluid equals that of the glass prism, then n 1 5 n 2 , and Snell’s law (n 1 sin u 1 5 n 2 sin u 2 ) reduces to sin u 1 5 sin u 2 . Therefore, the angle of refraction equals the angle of incidence, and no bending of the light occurs. 26.6 | Lenses 661 After reflection from the back surface of the droplet, the different colors are again refracted as they reenter the air.

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  Electromagnetism

  Maxwell Equations, Wave Propagation and Emission

  • Tamer Becherrawy(Author)
  • 2013(Publication Date)
  • Wiley-ISTE

    (Publisher)

  S p. The envelope Σ of these wavelets is the wavefront later. It is not necessary for the sources to be material ones; the principle holds even in the case of the propagation of light in vacuum. From this general principle, it is possible to deduce the following laws of reflection and refraction, which are verified experimentally:

  − The direction of propagation of the incident wave, that of the reflected wave, and that of the refracted wave lie in the same plane containing the normal to the interface at each point of incidence.

  − The angle of reflection θ′ (between the direction of propagation of the reflected wave and the normal Oz ′ to ) is equal to the angle of incidence θ (between the direction of propagation of the incident wave and the normal Oz to ). The angle of refraction θ" (between the direction of propagation of the refracted wave and the normal Oz to ) is related to θ by Snell’s law

  [11.1 ]

  n 1 =c/v 1 and n 2 =c/v 2 are the indices of refraction of mediums (1) and (2), where c is the speed of propagation in a medium of reference (the vacuum in the case of electromagnetic waves).

  Equation [11.1 ] determines the angle of refraction θ", if (n 1 /n 2 ) sin θ < 1. This condition can be always satisfied if n 1 < n 2 . In the case n 1 > n 2 , we must have sin θ < n 1 /n 2 . Thus, θ must be less than a critical angle (or limiting angle ) i L , given by

  [11.2]

  At the angle of incidence equal to i L the angle of refraction is θ" = 90°. If the angle of incidence θ is larger than i L , the wave undergoes total reflection .

  The wave theory enables us to establish the laws of reflection and refraction using the boundary conditions (or continuity equations ) at the interface . We have only to assume that these conditions are expressed as linear relations between the incident wave u , the reflected wave u ′, and the transmitted wave u ″, and their partial derivatives with respect to time or space coordinates. In the case of simple harmonic waves , the time derivative is simply iωu and the derivative with respect to x , for instance, is –ik x u . Thus, the boundary conditions are linear, of the general form au + bu ′ + cu "

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  General Physics Electromagnetism Optics

  • Pierluigi Zotto, Sergio Lo Russo, Paolo Sartori(Authors)
  • 2022(Publication Date)
  • Società Editrice Esculapio

    (Publisher)

  In all the other cases it is convenient to use the ray concept, which assumes that a radiation is coincident with a segment which shares the propagation direction with the considered wave. Such an approximation originates the geometrical optics discipline; if the approximation is not valid or it is not used, the disci- pline is called wave optics. A mixture between the two approaches to optics occurs fre- quently while describing optical phenomena, but distinguishing which of the two to apply is quite easy. 16.2 Snell’s Laws A wave which strikes a surface is partially transmitted in the material, thus generating a refracted (or transmitted) wave, and it is partially reflected by it, thus generating a reflected wave. A progressive electromagnetic wave which propagates along axis x is described by the equation E = f kx − ω t ( ) . A vector k = k  u x , called wave-vector, whose magnitude is the wave number k and whose direction is the wave propagation direction, can be introduced. Any point P of a wavefront is identified with respect to the origin of a reference system by 261 16 its position vector  r = x  u x + y  u y + z  u z , whose scalar product with wave-vector  k gives  k i  r = kx and therefore the wave can be described by the function E = f  k i  r − ω t ( ) , which allows the generalisation of the wave function to a wave which propagates in any direction with respect to an arbitrary reference system. A wave-vector represents the propa- gation direction of a wave, that is the direction of its light ray in geometrical optics. When a wave strikes a surface, an incident wave Ei, a reflected wave Er and a refracted (or transmitted) wave Et are simultaneously present in any point P struck by the beam.

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  Physics for Students of Science and Engineering

  • A. L. Stanford, J. M. Tanner(Authors)
  • 2014(Publication Date)
  • Academic Press

    (Publisher)

  to x by L 1 2 = x 2 + a 2 ; L 2 2 = (d − x) 2 + b 2 so 2 L 1 d L 1 d x = 2 x 2 L 2 d L 2 d x = − 2 (d − x) d L 1 d x = x L 1 = sin θ 1 d L 2 d x = − (d − x) L 2 = −[--=PLGO. -SEPARATOR=--]sin θ 2 Then because d t/ d x = 0, it follows that n 1 sin θ 1 = n 2 sin θ 2 (Snell ’ s law) where θ 1 and θ 2 are the angles of incidence and refraction, respectively. Total Internal Reflection When light is incident on an interface between two media, part of the light energy is, in general, reflected back into the medium of the incident ray and part is refracted and transmitted into the other medium. Figure 20.16(a) shows a ray being reflected and refracted at an interface between media. Experiment shows that the path of a refracted ray is reversible: If, as in Figure 20.16(b), the ray in the lower medium is reversed so that it is incident on the interface at the same angle θ 2 as in Figure 20.16(a), that ray is refracted at the angle θ 1. In both Figures 20.16(a) and 20.16(b), the reflected rays obey the law of reflection. Figure 20.16 Illustrating that the path of a light ray refracted at an interface between media is reversible. Let us now consider certain conditions under which Snell’s law cannot apply. Suppose, as in Figure 20.17, a light ray is refracted at an interface separating media having indices of refraction n 1 and n 2 > n 1. If θ 2 is the angle of incidence and θ 1 is the angle of refraction, we may write Snell’s law as Figure 20.17 A light ray refracted at an interface between media having indices of refraction n 1 and n 2 > n 1. sin θ 1 = n 1 n 2 sin θ 2 (20-21) (20-21) Because the sine of an angle must be equal to or less than 1, Equation (20-21) cannot be satisfied if its right-hand side has a value greater than 1. When θ 2 has the value θ c, called the critical angle, the angle of refraction θ 1 is equal to 90° so that sinθ 1 = 1

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  Elements of Modern X-ray Physics

  • Jens Als-Nielsen, Des McMorrow(Authors)
  • 2011(Publication Date)
  • Wiley

    (Publisher)

  Chapter 3 Refraction and reflection from interfaces

  A ray of light propagating in air changes direction when it enters glass, water or other transparent materials. This is the basis for the classical optics of lenses. Quantitatively, the phenomenon is described by Snell’s law. For visible wavelengths the refractive index n of most transparent materials has a value in the range between 1.2 and 2. The refractive index depends on the frequency ω of the light, so that blue light is refracted more than red light, etc.

  The index of refraction for electromagnetic waves displays resonant behaviour at frequencies corresponding to electronic transitions in atoms and molecules. On the low frequency side of a resonance, n increases with ω, and this is known as normal dispersion. Immediately above the resonance frequency it decreases, and as more and more resonances are passed, the magnitude of the index of refraction decreases. X-ray frequencies are usually higher than all transition frequencies, perhaps with the exception of those involving the inner K- or maybe L-shell electrons. As a result in the X-ray region n turns out to be less than unity. (See Fig. 1.8 and accompanying discussion.) This reflects the phase shift of π in the Thomson scattering of X-rays, as we shall see. Moreover, it leads to the phenomenon of total external reflection from a flat, sharp interface: for incident glancing angles α below a certain critical angle αc the ray will no longer penetrate into the material but will be totally reflected from it. The deviation of n from unity is tiny, so the critical angle is small. The reader might wonder how n can be less than unity, since the velocity in the material is c /n , and this would seem to imply that the speed of light is higher in the material than in vacuum. However, c /n is the phase velocity, not the group velocity. The latter, evaluated as d ω/d k, is indeed less than c

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  Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2021(Publication Date)
  • Wiley

    (Publisher)

  A ray bends away from the normal when it travels from a medium with a larger refractive index into a medium with a smaller refrac- tive index. When the ray leaves the prism, it again bends upward, which is toward the normal at the point of exit. A ray bends toward the normal when traveling from a smaller toward a larger refrac- tive index. Thus, thesituationin Figure26.19acouldariseifthe prismwereimmersedinafluid,suchascarbondisulfide,that hasalargerrefractiveindexthandoesglass (see Table 26.1). We have seen in Figures 26.18a and 26.19a that a glass prism can bend a ray of light either downward or upward, depend- ing on whether the surrounding fluid has a smaller or larger index of refraction than the glass. It is logical to conclude, then, that a prismwillnotbendarayatall,neitherupnordown,ifthesur- roundingfluidhasthesameindexofrefractionastheglass—a condition known as index matching. This is exactly what is happen- ing in Figure 26.19b, where the ray proceeds straight through the prism as if the prism were not even there. If the index of refraction of the surrounding fluid equals that of the glass prism, then n 1 = n 2 , and Snell’s law (n 1 sin θ 1 = n 2 sin θ 2 ) reduces to sin θ 1 = sin θ 2 . Therefore, the angle of refraction equals the angle of incidence, and no bending of the light occurs. Related Homework: Check Your Understanding 16 (a) (b) FIGURE 26.19 A ray of light passes through identical prisms, each surrounded by a different fluid. The ray of light is (a) refracted upward and (b) not refracted at all. THE PHYSICS OF . . . rainbows. Another example of dispersion occurs in rain- bows, in which refraction by water droplets gives rise to the colors. You can often see a rain- bow just as a storm is leaving, if you look at the departing rain with the sun at your back.

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  Physics

  • John D. Cutnell, Kenneth W. Johnson, David Young, Shane Stadler(Authors)
  • 2018(Publication Date)
  • Wiley

    (Publisher)

  How can the situations illustrated in Figure 26.19 arise? Reasoning and Solution Snell’s law of refraction includes the refrac- tive indices of both materials on either side of an interface. With this in mind, we note that the ray bends upward, or away from the normal, as it enters the prism in Figure 26.19a. A ray bends away from the normal 26.5 The Dispersion of Light: Prisms and Rainbows 747 THE PHYSICS OF . . . rainbows. Another example of dispersion occurs in rainbows, in which refraction by water droplets gives rise to the colors. You can often see a rainbow just as a storm is leaving, if you look at the departing rain with the sun at your back. When light from the sun enters a spherical raindrop, as in Figure 26.20, light of each color is refracted or bent by an amount that depends on the refractive index of water for that wavelength. After reflection from the back surface of the droplet, the different colors are again refracted as they reenter the air. Although each droplet disperses the light into its full spectrum of colors, the observer in Figure 26.21a sees only one color of light coming from any given droplet, since only one color travels in the right direction to reach the observer’s eyes. However, all colors are visible in a rainbow (see Figure 26.21b) because each color originates from different droplets at different angles of elevation. Water droplet Sunlight Violet Red FIGURE 26.20 When sunlight emerges from a water droplet, the light is dispersed into its constituent colors, of which only two are shown. FIGURE 26.21 (a) The different colors seen in a rainbow originate from water droplets at different angles of elevation. (b) A rock climber beneath a rainbow. Sun Violet Violet Red (a) Red Red Violet (b) ©ACE STOCK LIMITED/Alamy when it travels from a medium with a larger refractive index into a medium with a smaller refractive index. When the ray leaves the prism, it again bends upward, which is toward the normal at the point of exit.

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