Light: Geometric Optics

Transcription

1 Light: Geometric Optics

2 The Ray Model of Light Light very often travels in straight lines. We represent light using rays, which are straight lines emanating from an object. This is an idealization, but is very useful for geometric optics.

3 When any wave meets a boundary there is a reflected and transmitted wave. A ray of light will obey the law of reflection: Angle of incidence equals angle of reflection. The transmitted wave is usually refracted (changes direction) unless the light ray were head on. The frequencies of the reflected and transmitted waves must equal that of the incident wave.

4 angle of! incidence angle of! reflection Mediium 1 Meidum 2 Boundary Boundary angle of! refraction

5 Example: What is the frequency of a light wave in vacuum that has a wavelength of 465 nm? Solution: υ = f λ or f = υ λ f= 3x10 8 m/s/ 465x10-9 m= 6.45 x10 14 Hz

6 Index of Refraction Refraction In general, light slows somewhat when traveling through a medium. The index of refraction of the medium is the ratio of the speed of light in vacuum to the speed of light in the medium: Speed of light (c) in vacuum: 3.00x10 8 m/s

7 Example: If a material has an index of refraction of 1.35, how fast does light travel in this material? Solution: υ = c n V= 3.00x10 8 m/s /1.35 = 2.22x10 8 m/s

8 Light changes direction when crossing a boundary from one medium to another. This is called refraction, and the angle the outgoing ray makes with the normal is called the angle of refraction.

9 Apparent Depth Refraction is what makes objects halfsubmerged in water look odd.

10 The angle of refraction depends on the indices of refraction, and is given by Snell s law: See derivation Other useful relationships given that the frequency of the refracted ray is equal to the incident ray. υ 1 λ 1 = υ 2 λ 2

11 Example: A light wave in air with a frequency of 4.62x10 14 Hz enters a piece of glass (n= 1.50) with an angle of incidence of a) What is the angle of refraction? Sketch a diagram. b) What is the wavelength ( in nm) of the light in the glass?

12 a) Solution: sinθ 2 = n 1 n 2 sinθ 1 = (1/1.5) x (sin 30)= ϴ 2 = sin -1 (0.3333)= n 2 = 1.50 glass air n 1 = ϴ 2 =? b) υ = c n υ = f λ or λ= υ f = c nf = (3x10 8 )/[(1.5)(4.62x10 14 )] = 4.33x10-7 m = 433 nm

13 Total Internal Reflection(TIR) If light passes into a medium with a smaller index of refraction, the angle of refraction is larger. There is an angle of incidence for which the angle of refraction will be 90 ; this occurs at an angle of incidence called the critical angle. At the critical angle and beyond, there is only a reflected ray and not transmitted one. The light has been totally internally reflected.

14 If the angle of incidence is larger than this, no transmission occurs. This is called total internal reflection.

15 Binoculars often use total internal reflection; this gives true 100% reflection, which even the best mirror cannot do.

16 Total internal reflection is also the principle behind fiber optics. Light will be transmitted along the fiber even if it is not straight. An image can be formed using multiple small fibers.

17 Example: A light ray travelling in glass (n=1.50) reflects at a critical angle of What is the index of refraction of the other medium? Solution: n 1 sinθ c = n 2 sin90 0 n 2 =? n 2 = n 1 sinθ C = 1.5 sin ( )= 1.33 n 1 = 1.50 ϴc

18 Reflection; Image Formation by a Plane Mirror Law of reflection: the angle of reflection (that the ray makes with the normal to a surface) equals the angle of incidence.

19 When light reflects from a rough surface, the law of reflection still holds, but the angle of incidence varies. This is called diffuse reflection.

20 With diffuse reflection, your eye sees reflected light at all angles. With specular reflection (from a mirror), your eye must be in the correct position.

21 What you see when you look into a plane (flat) mirror is an image, which appears to be behind the mirror. Two light rays from a point on the object are required to find the corresponding point on the image. The object is the original source of light. The image is formed by rays leaving the object Object Image

22 This is called a virtual image, as the light rays do not converge; they diverge on reflection in this case. The distance of the image from the mirror is equal to the distance of the object from the mirror.

23 Formation of Images by Spherical Mirrors Spherical mirrors are shaped like sections of a sphere, and may be reflective on either the inside (concave) or outside (convex).

24 Rays coming from a faraway object are effectively parallel.

25 Parallel rays striking a spherical mirror do not all converge at exactly the same place if the curvature of the mirror is large; this is called spherical aberration.

26 If the curvature is small, the focus is much more precise; the focal point is where the parallel rays converge.

27 Using geometry, we find that the focal length is half the radius of curvature: Spherical aberration can be avoided by using a parabolic reflector; these are more difficult and expensive to make, and so are used only when necessary, such as in research telescopes.

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29 The intersection of these three rays gives the position of the image of that point on the object. To get a full image, we can do the same with other points (two points suffice for many purposes).

30 Geometrically, we can derive an equation that relates the object distance, image distance, and focal length of the mirror:

31 We can also find the magnification (ratio of image height to object height). The negative sign indicates that the image is inverted. This object is between the center of curvature and the focal point, and its image is larger, inverted, and real.

32 Notes on Sign Conventions If the di is positive the image is real. This means the light rays converge and the image could be projected on a screen. If di is negative the image is virtual. The light rays diverge and appear to originate from a common point. For single optical systems do is always positive. f > 0 for concave (converging ) spherical mirror f<0 for convex (diverging) spherical mirror

33 If an object is outside the center of curvature of a concave mirror, its image will be inverted, smaller, and real.

34 If an object is inside the focal point, its image will be upright, larger, and virtual.

35 Example: A 5.00 cm tall object is 15 cm from a concave mirror with radius of curvature of 20.0 cm a) What is the focal length? b) Calculate the image distance c) Sketch the ray diagram d) How tall is the image? c f

36 a) f= 20/2 = 10 cm b) 1 d i = 1 f 1 d o = = = 1 30 di= 30 cm c) c f d) m= -d i /d o = -30 cm / 15 cm= -2 h i = mh o =-2 ( 5 cm)= -10 cm

37 Example: An object when placed 4.00 cm from a concave spherical mirror produced a virtual image three times larger. a) What is the focal length of the mirror? b) Sketch the ray diagram

38 Solution: a) do= 4.00 cm d i d o = - 3 di=(-3)( 4 cm)= -12 cm 1 f = 1 d o + 1 d i = = = 2 12 = 1 6 f= 6.00 cm b) f

39 For a convex mirror, the image is always virtual, upright, and smaller. Parallel light rays diverge on reflection from surface.

40 Example: A 5.00 cm tall object was 10.0 cm from a spherical mirror which produced an upright virtual image half its size. a) What kind of mirror must this be? b) Draw the ray diagram. c) If an object were placed 15.0 cm from the mirror, what would be the location of the image?

41 Solution: a) It must be a convex spherical mirror as it is the only spherical mirror that produces a virtual image of reduced size. b) object image c) Determine the focal length first then calculate the image distance using d 0 = 15 cm

42 Problem Solving: Spherical Mirrors 1. Draw a ray diagram; the image is where the rays intersect. 2. Apply the mirror and magnification equations. 3. Sign conventions: if the object, image, or focal point is on the reflective side of the mirror, its distance is positive, and negative otherwise. Magnification is positive if image is upright, negative otherwise. 4. Check that your solution agrees with the ray diagram.

43 Thin lenses are those whose thickness is small compared to their radius of curvature. They may be either converging (a) or diverging (b). Thin Lenses

44 Parallel rays are brought to a focus by a converging (convex) lens (one that is thicker in the center than it is at the edge).

45 A diverging (concave) lens (thicker at the edge than in the center) make parallel light diverge; the focal point is that point where the diverging rays would converge if projected back.

46 The power of a lens is the inverse of its focal length. Lens power is measured in diopters, D. 1 D = 1 m -1

47 Ray Tracing Ray tracing for thin lenses is similar to that for mirrors. We have three key rays: 1. This ray comes in parallel to the axis and exits through the focal point. 2. This ray comes in through the focal point and exits parallel to the axis. 3. This ray goes through the center of the lens and is undeflected.

48 The Thin Lens Equation The thin lens equation is the same as the mirror equation:

49 The sign conventions for lenses 1. The focal length is positive for converging (convex) lenses and negative for diverging (concave). 2. The object distance is positive when the object is on the same side as the light entering the lens (not an issue except in compound systems); otherwise it is negative. 3. The image distance is positive if the image is on the opposite side from the light entering the lens; otherwise it is negative. 4. The height of the image is positive if the image is upright and negative otherwise.

50 The magnification formula is also the same as that for a mirror: The power of a lens is positive if it is converging and negative if it is diverging.

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52 Example: A 4.00 cm high object is 20 cm from a convex lens of focal length + 12 cm. a) Determine the position of the image b) Determine the height of the image c) Draw the ray diagram

53 Solution: a) 1 = 1 d i f 1 = 1 d o = = 1 30 di= 30 cm b) hi/ho = -di/do, hi = (-30/20)(4 cm)= - 6 cm c) object F F image

54 Example: An object is 5.00 cm from a convex lens with a focal length of cm a) What is the location of the image b) What is its magnification c) Draw the ray diagram d) Describe the image (eg. Size, orientation, kind etc)

55 Solution: a) 1 d i = 1 f 1 d o = di= -15 cm b) m = d i d o = - (-15 cm)/(5 cm) = 3.0 cm c) virtual image object d) The image is virtual, upright, and enlarged

56 For a diverging lens, we can use the same three rays; the image is upright and virtual.

57 Example: A 9 cm tall object is 27 cm in front of a concave lens with a focal length -18 cm. a) Determine the image distance b) What is the height of the image c) Draw the ray diagram

58 Solution: a) 1 d i = 1 f 1 d o = di= cm b) hi/ho = -di/do, hi = (10.8/27)(9 cm)= 3.6 cm C) object image

59 Harder lens problem: An object is a fixed distance ( 25 cm) from a screen. Where should a convex lens (f= 4.00 cm) be placed (x) so the a real image is focussed on the screen? lens screen x=? object 25 cm

60 Solution: d i = 25 d o 1 d o d o = 1 4 = 25 d o + d o d o (25 d o ) d o 2 25d o +100 = 0 do =x = 5.00 cm and 20 cm

61 The Visible Spectrum and Dispersion Wavelengths of visible light: 400 nm to 750 nm Shorter wavelengths are ultraviolet; longer are infrared

62 The index of refraction of a material varies somewhat with the wavelength of the light.

63 This variation in refractive index is why a prism will split visible light into a rainbow of colors.

64 Actual rainbows are created by dispersion in tiny drops of water.

65 Polarization Light is polarized when its electric fields oscillate in a single plane, rather than in any direction perpendicular to the direction of propagation.

66 Polarized light will not be transmitted through a polarized film whose axis is perpendicular to the polarization direction.

67 This means that if initially unpolarized light passes through crossed polarizers, no light will get through the second one.

68 Light is also partially polarized after reflecting from a nonmetallic surface. At a special angle, called the polarizing angle or Brewster s angle, the polarization is 100%.

69 Why are Skies blue and Sunset Red Sunlight (made up of all colours) reaches Earth's atmosphere and is scattered in all directions by all the gases and particles in the air. Blue light is scattered in all directions by the tiny molecules of air in Earth's atmosphere. Blue is scattered more than other colors because it travels as shorter, smaller waves. This is why we see a blue sky most of the time. Courtesy of

70 As the Sun gets lower in the sky, its light is passing through more of the atmosphere to reach you. Even more of the blue light is scattered, allowing the reds and yellows to pass straight through to your eyes.