Refraction at Plane Surfaces, Lecture notes of Chemistry

Download Refraction at Plane Surfaces and more Lecture notes Chemistry in PDF only on Docsity! REFRACTION AT PLANE SURFACES CHAPTER THREE REFRACTION AT PLANE SURFACES Refraction When light is incident on a boundary between two media of different optical densities two physically meaningful phenomena occur. We may have reflection of light at the boundary or we may have refraction of light at the boundary. It is the latter of the two physical phenomena, which we are interested in; in this part of the book. In Figure 3.1, an incident ray of light on the boundary between two optically different media will undergo refraction in addition to reflection. According to the laws of reflection the angle of reflection is the same as the angle of reflection. The relationship between the angle of incidence i and the angle of refraction r, will be later defined by Snell’s law but first let’s look at the basic laws governing the refraction. From here, we can define refraction as “the bending of light rays as it moves from one medium to another of different optical density”. There are three basic laws of refraction (a) The incident ray, the refracted ray and the normal to the surface (or the boundary) all lie in the same plane. 49 Reflected rayIncident ray Refracted ray Figure 3.1 When light strikes a boundary it undergoes reflection in addition to refraction REFRACTION AT PLANE SURFACES (b) The path of a ray refracted at the interface between two media is exactly reversible. (c) The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. (This is also known as Snell’s law) These laws are easily demonstrated by observation and experiment. However, it is of much more importance in a practical sense, to understand and predict the degree of bending. Index of Refraction We now know that light bends but in which direction does it bend. Is it away from the normal to the boundary or towards the normal? From Figure 3.2, light when passing from a less dense medium to a denser medium will upon refraction bend towards the normal N and when passing from a denser medium to a less dense medium will upon refraction bend away from the normal N. One physical parameter which can give a measure or degree of the bending after refraction is the index of refraction or simply the refractive index. The refractive index n of a particular material is the ratio of the speed of light in vacuum or free space to the velocity of light through the material. The refractive index (or the index of refraction) is a dimensionless or unitless quantity. For example, . Note that the values of refractive indices given for water and glass apply to yellow light of wavelength 589nm (1nm =1 nano meter = ). Figure 3.3 shows the refraction of light at the boundary between two media. The speed of light in material substances is different for different wavelengths. This effect is known as dispersion. 50 REFRACTION AT PLANE SURFACES Assuming the velocity v2 in the second medium is smaller than the velocity v1 in the first medium, the distance AC will be shorter than the distance BD. These lengths are given by It can be shown from geometry that angle BAD is equal to θ1 and that angle ADC is equal to θ2, as indicated in Figure 3.4. The line AD forms a hypotenuse that is common to the two triangles ADB and ADC. From Figure 3.4; 3.2a 3.2b Dividing Equation 3.2a by Equation 3.2b, we obtain 3.3 The ratio of the sine of the angle of incidence to the sine of the angle of the refraction is equal to the ratio of the speed of light in the incident medium to speed of light in the refracting medium. This rule was discovered by a Dutch astronomer Willebrord Snell and is called Snell’s law in his honor. An alternative form for the law can be obtained by expressing the velocities v1 and v2 in terms of the indices of refraction for the two media. By using Equation 3.1, we have 3.4 Putting Equation 3.4 in Equation 3.3 we get 3.5 53 REFRACTION AT PLANE SURFACES Since the sine of an angle increases as the angle increases, we see that an increase in the index of refraction results in a decrease in the angle and vice versa. Alternative Notation for Refractive Indices In this book the notation for refractive index, n can be used interchangeably with η (eta) in other texts. Please do not be confused they represent the same thing. The refractive index of material 2 with respect to material 1 can written mathematically as 3.6 where is the absolute refractive index of material 1 and the absolute refractive index of material 2. Refractive Index and Wavelength When light passes from a less dense medium to a denser medium the light slows down. The general question is what happens to the wave length of light entering a new medium. In Figure 3.4, light traveling in air at a velocity c encounters a medium through which it travels at the reduced speed vm. Upon entering to air again, it will travel at the speed c of light in air. The frequency is the same inside the medium and in air. The velocity is related to the frequency and wavelength by 3.7a 3.7b where c and vm are the speeds in air and inside the medium and λa and λm are the respective wavelengths. Since the velocity decreases inside the medium, the wavelength inside the medium must decrease 54 REFRACTION AT PLANE SURFACES proportionally for the frequency to remain constant. Dividing Equation 3.7a by Equation 3.7b yields 3.8 If we substitute , we obtain 3.9 Therefore the wavelength λm inside the medium is given by 3.10 where nm is the refractive index of the medium and λa is the wavelength of light in air. Generally we can write Snell’s law for light traveling from medium 1 to medium 2 as 3.11 Example 3.1 A beam of light enters the flat surface of diamond from air at an angle of 30° from the normal. The angle of refraction in the diamond is measured to be 12° from the normal. What is the refractive index of this diamond? Interpret your answer physically. Solution: Let medium 1 be air and medium 2 be diamond i.e. . Using Snell’s law (Equation 3.11), we can find n2 which is the refractive index of diamond. is the refractive index for air. 55 REFRACTION AT PLANE SURFACES Refraction causes an object immersed in a liquid of higher refractive index to appear closer to the surface than it actually is. This shallowing effect is illustrated in Figure 3.6. The object O appears to be at I because of refraction of light from the object. The apparent depth is denoted by v and the real depth by u. Snell’s law applied at the surface gives 3.14 From Figure 3.6, we can relate the ratio of the refractive indices to the actual and apparent depth by the following: we note that and also that Using these in Equation 3.14, we obtain 58 θ 2 θ 1 n 2 n 1 N A O D v u θ 2 θ 1 B I d Figure 3.6 Relation between real depth and apparent depth REFRACTION AT PLANE SURFACES 3.15 If we restrict ourselves to rays that are nearly vertical, the angles θ1 and θ2 will be small, so that the following approximations apply where u and v are the real and apparent depths respectively. Applying these approximations to Equation 3.15, we can then write 3.16 3.17 From Figure 3.6, real depth u is equal to apparent depth v plus the deviation (or lateral displacement) d which is given by 3.18 where d is the displacement. But from Equation 3.17, . If medium 1 is glass ( ) and medium 2 is air ( ) then we write 3.19 substituting v in Equation 3.18 we get therefore the displacement d is given by 3.20 where nm is the refractive index of medium 2 (material) and u is the real depth. We can use this displacement method to find the refractive index of solid transparent materials and substances. 59 REFRACTION AT PLANE SURFACES Example 3.4 A coin rests on the bottom of a container filled with water (nw=1.33). The apparent depth (distance) of the coin from the surface is 9cm. How deep is the container? Solution: We have as the refractive indices of water and air respectively. The actual depth u is found by solving for u in Equation 3.19. The apparent depth is approximately three-quarters of the actual depth. Example 3.5 When light of one wavelength from air hits a smooth piece of glass at an angle, which physical phenomena occur? Solution: The answer is reflection and refraction. Reflecting prism Totally reflecting prisms are used in a variety of optical instruments such as periscopes, prismatic binoculars, cameras and projection 60 Reflected rayIncident ray Refracted ray Figure 3.7 Both reflection and refraction occur at boundaries between two media REFRACTION AT PLANE SURFACES In : This is from geometry where the sum of all angles in a triangle is equal to 180°. thus 3.25 Equations 3.24 and 3.25 are the general equations that hold for refraction in prisms. Minimum Deviation of a prism Experiment and theory show that the minimum deviation, Dmin of the light the passing through a prism occurs when the rays pass symmetrically through the prism. When this happens, the internal ray XY is parallel to the base of the prism (QR) and therefore and . Substituting these into Equations 3.24 and 3.25 we have 3.26 3.27 Similarly 3.28 Relationship between A, Dmin and n We can establish the relationship between A, Dmin and n by adding Equations 3.27 and 3.28 to get 3.29 63 REFRACTION AT PLANE SURFACES 3.30 From Equation 3.28 3.31 From the Snell’s law 3.32 3.33 but 3.34 where is the refractive index of glass with respect to air. Experiment shows that as the incident angle i, is increased from zero, the deviation D begins to decrease continuously to some minimum value which is the minimum deviation and then increases further to 90°. Figure 3.9 shows a graph of deviation plotted against the angle of incidence i. In Figure 3.9 the deviation D will decrease from a maximum deviation to a minimum Dmin and then increase again to Dmax. 64 REFRACTION AT PLANE SURFACES Grazing angles When the angle of incidence on the face of the prism is 90, the ray is said to be having a grazing incidence on the face of the prism and it emerges at the other face making an angle of i2 to the normal. 65 i D minD maxD 90Figure 3.9 Variation of deviation, D, with angle of incidence i, for refraction through prisms. Figure 3.10 Light incident on a prism with a grazing incidence angle of with the final ray after refraction through the prism emerging with an angle of 2r1r 90 A n 2i grazing incident ray emergent ray REFRACTION AT PLANE SURFACES reversed position the coloured beams between the two prisms emerge in the same direction and white light is again obtained. There are three states of purity of a spectrum:  Impure Spectrum: There are many shades of each colour, each shade merging gradually into the next.  Fairly pure Spectrum: In between pure and impure.  Pure Spectrum: Distinct separation of colours. Refractive Index and the colour of light Refractive index depends not only on the medium but also on the type of light used. Colour of Opaque objects When white light falls on a body and the body reflects all the colours in the white light, the body appears white. If however the body reflects some of the colours while absorbing the others, then the body appears coloured. The colour of an opaque body depends on the kind of light falling on it as well as the colours of light which it absorbs or reflects. Exercises 3.1 lights in vacuum is incident on the surface of a glass slab. In the vacuum the beam makes an angle of 32° with the normal to the surface, while in the glass it makes an angle of 21° with the normal. What is the index of refraction of the glass? Ans.: 1.48. 3.2 A layer of benzene (index of refraction = 1.50) floats on water. If the angle of incidence of the light entering the benzene from air is 60°, 68 REFRACTION AT PLANE SURFACES what is the angle the light makes with the vertical in the benzene and in the water? Ans.: 35° in benzene and 41° in water. 3.3 A glass dish with a plane parallel bottom and refractive index1.51 is half-filled with water. Then carbon disulfide is poured on top of the water. Finally, a flat cover of the same type of glass is placed on top of the dish. A beam of light making an angle of 50° with the vertical is incident on the horizontal cover. Find the angles which the beam makes with the vertical as it passes through glass, carbon disulfide, water, glass and air. Ans. θglass = 30.5°, θcarbon disulfide = 28°, θwater = 35.1°, θ′glass = 30.5°, θair = 50°. 3.4 When a fish looks up at the surface of a perfectly smooth lake, the surface appears dark except inside a circular area directly above it. Calculate the angle φ that this illuminated region extends. Ans.: φ = 97.2°. 3.5 (a) What is the speed of a crystalline quartz? (b) The speed of light in zircon is . What is the index of refraction of zircon? Take the index of refraction of quartz to be 1.553. Ans.: (a) v = (b) n = 1.97. 3.6 Light having a free-space wavelength of passes from vacuum into diamond ( ). Under ordinary circumstances the frequency is unaltered as light traverses different substances. Assuming this to be the case, compute the wave’s speed and wavelength in the diamond. Ans.: , . 3.7 At what angle must a ray of light incident on acetone to be refracted into the liquid at 25°? ( ) Ans.: 35°. 69 REFRACTION AT PLANE SURFACES 3.8 What is the speed of light in water? Find the angle of refraction of light incident on a water surface at angle of 48° to the normal. (n = 1.333). Ans.: , . 3.9 To some extent the brilliance of diamonds is attributable to total internal reflection. Calculate the critical angle for a diamond-air surface. (ndiamond = 2.42). Ans.: . 3.10 If the speed of light in ice is . What is its index of refraction? What is the critical angle of the incidence for light going from ice to air? Ans.: nice = 1.304, . 70