Unit 3 - Foundations of Waves Chapter 6 - Light, Mirrors, and Lenses Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 1 / 57
Section 6.1 - The Behaviour of Light History of Light Plato (428 BCE - 348 BCE) thought that light consisted of streams of particles emitted from the eyes. Aristotle (384 BCE - 322 BCE) thought the streams were emitted in the other direction. Newton (1642-1727) postulated that light consists of streams of particles. Huygens (1629-1695) and others supported the idea that light does not consist of particles but rather of waves. Einstein, Planck, Broglie, and Bohr (20th century) proposed and confirmed that all particles also have a wave nature (and vice versa). This is called wave-particle duality. Watch: The History of Light Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 2 / 57
Sources of Light Objects can be seen only when a source of light is present. There is a difference between luminous bodies, such as stars, incandescent and fluorescent lamps, which emit light, and illuminated bodies such as the trees, grass and moon, which reflect light. Most visible objects are illuminated by a luminous source and are seen by light, which they reflect. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 3 / 57
Man Made Light Solids give off light when heated. When heated, a solid s electrons do not move in exactly the same way. The resulting light has a whole range of frequencies, which extend from ultraviolet to infrared, which our eyes cannot see. These heated materials are known as incandescent substances. Gasses such as neon and sodium give off light when electric currents are passed through them or when they are heated, but unlike a solid, a gas gives off light at a unique frequency. Neon is red and sodium is orange. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 4 / 57
Fluorescent lamps produce light through mercury vapours, which produces ultraviolet rays. These rays strike the interior of the tube or bulb, which is coated with chemicals called phosphors. These chemicals give off visible light when bombarded by ultraviolet rays. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 5 / 57
Natural Light Stars produce light by the internal nuclear reactions occurring in the centre of each star. The temperature of the star s surface determines the colour of the light. Very hot stars are white and cool stars are red. Other sources of natural light are lightning, produced from electric charges rushing through the atmosphere between clouds and the ground, Aurora Borealis, produced when air molecules interact with charged particles emitted from the Sun that are caught in Earth s magnetic field, and bioluminescent animals such as fireflies and electric eels whose light is produced by chemical reactions. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 6 / 57
Transmission and Absorption of Light There are three different classes of absorption: Transparent - Materials such as glass, quartz, and air allow the passage of light in straight lines. Translucent - Materials such as paper and frosted glass allow light to pass through but in different directions so that one cannot see objects through them. Opaque - Materials such as wood, brick and metals allow no light to pass through. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 7 / 57
Dispersion When light is passed through transparent crystals brilliant colours of light are produced. At first this phenomenon was attributed to the crystal but Newton was able to show that these colours where already present in light. If a beam of white light is allowed to pass through a prism the components of light are refracted at various amounts to produce a series of colours called a spectrum. This is called dispersion. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 8 / 57
The Light Spectrum Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 9 / 57
Why is the Sky Blue? The scattering of light is why the sky is blue during the day and red during the sunset. Scattering (α) is inversely proportional to the 4th power of the wavelength (α = 1 4λ ). 1 The short waves of violet light are scattered easily compared to the longer red waves. Therefore when light enters the earth s atmosphere the frequency that is scattered most is that of blue. 2 At night since the light must travel a greater distance through the atmosphere because of the angle light enters the atmosphere the red waves pass through the atmosphere while the other colours are completely scattered and the sun becomes progressively redder as it sets. (Video) Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 10 / 57
How does light travel? Light travels in straight lines. This is evident by observing the shadows produced by objects. If the light source is a point then the shadow produced by an image will be dark and sharp since no light is reflected. It is important to note that directly around the shadow is a region where the shadow is slightly fuzzy. This is caused by the diffraction of light. When light travels close to the edge of any object, or opening, it is bent in its path and travels in a new direction. Recall: This is diffraction. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 11 / 57
If the light source is not a point then two types of shadows will be created. A region where no light is reflected called the umbra and a region where partial reflection occurs which is called the penumbra. Solar and Lunar eclipses follow the pattern above. The following diagram shows how the lunar eclipse will appear from different parts of the earth. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 12 / 57
Application: Images in a Pinhole Camera A pinhole camera is effectively a light-proof box with a small hole in one side. Light from a scene passes through this single point and projects an inverted image on the opposite side of the box. Key fact: If a source of light passes through a tiny hole than the image produced is inverted. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 13 / 57
The relationship between the object and the image is given by the formula: height of image distance of image from pinhole = height of object distance of object from pinhole Symbolically, we represent this as: h 1 h 2 = d 1 d 2 Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 14 / 57
Example: Determine the height of an image produced by an object 12.2 m in height that is 19.7 m from a camera with a lens to film length of 38 mm. h 1 h 2 = d 1 d 2 h 1 =?; h 2 = 12.2 m d 1 = 38 mm = 0.038 m; d 2 = 19.7 m h 1 12.2 m = 0.038 m 19.7 m (12.2 m)(0.038 m) h 1 = 19.7 m h 1 = 0.024 m = h 1 = 24 mm Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 15 / 57
Speed of Light The speed of light can change depending on the medium that it is travelling through. Therefore the speed will differ in our atmosphere and space. The modern value for the speed of light in a vacuum is 299 792 400 m/s. For calculation purposes, c = 3.00 10 8 m/s. A light year is the distance light will travel in 1 year. = 1 light year = 9.46 10 15 m Time of flight techniques Cavity resonance techniques Watch: The Original Method Watch: Using a Microwave to Calculate the Speed of Light Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 16 / 57
Example: Light takes 1.28 s to travel from the moon to the Earth. What is the distance between them? Recall: v = d t Since we re interested in the movement of light in space: v = c = 3.00 10 8 m/s t = 1.28 s d = (v)( t) = (3.00 10 8 m/s)(1.28 s) = 3.84 10 8 m Since I didn t ask, you can leave the distance in this unit. If you wish, you can convert it to a more appropriate unit: 384 000 km Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 17 / 57
Practice: How much time would be required for a spaceship travelling at 60.0 10 4 km/h to reach the closet known star, Proxima Centauri, 4.3 light-years away? [See Section 6.1 Video] Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 18 / 57
Section 6.2 - Reflection of Light When light is reflected off of a surface, the type of reflection depends on the smoothness of the surface. Regular reflection occurs when reflecting bodies are smooth such as a mirror. Diffuse reflection occurs when the reflecting surface is light in colour but irregularly shaped. Reflex reflection occurs when light is reflected from a manufactured surface such as traffic signs. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 19 / 57
Laws of Reflection: 1 The angle of incidence is equal to the angle of reflection. 2 The incident ray, the reflected ray, and the normal to the reflecting surface lie in the same plane. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 20 / 57
Images in a Plane Mirror When you look into a mirror, you see an image of your face apparently located behind the mirror. If you move towards the mirror, your image will move closer to the mirror so that your image is always the same distance from the mirror as the object. If you raise your right hand the image on the mirror appears to be a left hand coming up to meet your right hand. Letters viewed in a plane mirror appear reversed horizontally but not vertically. This is called lateral inversion. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 21 / 57
Example: A student stands 3.0 m in front of a plane mirror. a) How far behind the mirror is his virtual image? Draw a diagram of the situation: Light that bounces off of the man needs to travel 3.0 m to hit the mirror, then is reflected 3.0 m to reach the man s eyes. Therefore, it appears to the man that his virtual image is 3.0 m behind the mirror. b) If he steps forward 1.0 m, what distance will separate him from his virtual image? Now he is 2.0 m from the mirror (and his virtual image appears 2.0 m behind the mirror). Therefore, the total distance is 4.0 m. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 22 / 57
Applications of Plane Mirrors Plane mirrors have many applications one of which is a looking glass i.e. mirror. Other applications are: Cameras Periscope Optical Lever Kaleidoscope See-through Mirrors Theatrical Effects Right Angle Images Watch: Magic with Mirrors Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 23 / 57
Reflection from Curved Surfaces Two types of curved mirrors used are convex or concave. 1 Concave refers to a reflecting surface that curves inwards. This type of mirror will make parallel light rays converge. 2 Convex refers to a reflecting surface that curves outwards. This type of mirror will make parallel light rays diverge. The centre of curvature (C) is the centre of a curved mirror. The radius (R) of curvature is any straight line drawn from C to the curved surface. The geometric centre of a curved mirror is called the vertex (V) and the line drawn through C and V is called the principal axis. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 24 / 57
Concave Mirrors Rays of light that are parallel to the principal axis that strike a concave mirror are reflected through a common point. This common point is called the focus. The focus is a point that lies on the principal axis. The distance from the vertex to the focus is called focal length (f). f = radius of curvature 2 Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 25 / 57
Rules for Rays in Curved Mirrors A ray parallel to the principal axis is reflected through the focus. A ray that passes through the focus is reflected parallel to the focal length. A ray that passes through point C is reflected back along the same path. You ll use at least two of these rules to determine the location and characteristics of an image given the starting position of some object in front of a curved mirror. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 26 / 57
The characteristics of an image formed in a concave mirror depend on the position of the object. There are essentially five cases: Object Position Image Position Size of Image Orientation of Image Type of Image Object distance is Image is between Smaller Inverted Real greater than C C and F Object is on C Image is on C Same size Inverted Real Object is between Image is greater Larger Inverted Real C and F than C Object is on F No image Object is in front Image is behind Larger Upright Virtual of F the mirror Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 27 / 57
Case 1: Object > C You ll find a copy of each of these diagrams (Cases 1-5) in the Chapter 6 Practice Problems package. You ll be expected to predict the location of the image. To draw these, follow the 3 laws outlined on slide 26. You need only use 2 of the laws to determine the position and characteristics of the image. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 28 / 57
Case 2: Object = C [See Section 6.2 Video] Try and predict the location and characteristics of the image that would be reflected. The solutions for the table above are found on slide 27. See the video for a stepby-step walkthrough. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 29 / 57
Case 3: F < Object < C [See Section 6.2 Video] Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 30 / 57
Case 4: Object = F [See Section 6.2 Video] Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 31 / 57
Case 5: Object < F [See Section 6.2 Video] Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 32 / 57
Convex Mirrors Regardless of the position the object, the image is always upright, small and virtual. Note: Virtual means that the reflected light rays never converge. This means you cannot capture the image on a screen; i.e. it doesn t really exist. So why do we see it? Our eyes will extrapolate where the light rays would have converged (see the dotted lines in the image below). Therefore, the image only exists in our brains. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 33 / 57
Section 6.3 - The Refraction of Light Light travels in lines only when the medium through which it travels remains the same. If it passes through a different medium, such as from air to water, the light rays will bend. This is called refraction. The index of refraction is the term used to compare the amount of refraction of light in mediums and is found using the following formula. Index of refraction of a medium = speed of light in a vacuum speed of light in the medium n = c v Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 34 / 57
Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 35 / 57
Example: The index of refraction of water is 1.33. Calculate the speed of light in water. n = c v n = 1.33 c = 3.00 10 8 m/s Note: c is a constant value. You will never need to solve for c. v =? v = c n = 3.00 108 m/s 1.33 = 2.26 10 8 m/s Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 36 / 57
Practice: What is the index of refraction of a solid in which the speed of light is 1.24 10 8 m/s? Identify the solid. Reference the table above. [See Section 6.3 Video] Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 37 / 57
Snell s Law This is used to determine the direction that light will travel into a specific medium. n = sin θ i sin θ R θ i - angle of incidence θ R - angle of refraction n - index of refraction Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 38 / 57
Examples: 1) A ray enters a medium of unknown index of refraction at 30. The angle of refraction is 24. Calculate the index of refraction to 2 s.f. n = sin θ i sin θ R θ i = 30 θ R = 24 n = sin 30 sin 24 = 1.2 Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 39 / 57
2) A ray of light enters water (n w = 1.33). If the angle of incidence is 30, what is the angle of refraction? Go to 2 s.f. n = sin θ i sin θ R θ i = 30 θ R =? n = 1.33 Solving for θ R will require us to use the sin 1 function (usually 2nd function sin on most calculators). sin θ R = sin θ i sin 30 = sin θ R = n 1.33 ( ) sin 30 θ R = sin 1 = θ R = 22 1.33 Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 40 / 57
Snell s law can also be used to determine the direction light will travel from one medium to a different medium. In this case the formula used is n 1 and n 2 correspond to the two different mediums of interest. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 41 / 57
Example: Light travels from glass into air. The angle of refraction in air is 60.0. What is the angle of incidence in glass if the index of refraction for glass is 1.52 and the index of refraction for air is 1.00? n 1 sin θ i = n 2 sin θ R 1 = glass; 2 = air n 1 = 1.52 n 2 = 1.00 θ i =? θ R = 60.0 (1.52)(sin θ i ) = (1.00)(sin 60.0 ) sin θ i = (1.00)(sin 60.0 ) 1.52 θ i = sin 1 ( (1.00)(sin 60.0 ) 1.52 ) = 34.7 Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 42 / 57
Practice: [See Section 6.3 Video] 1) Light travels from glass into water. The angle of incidence in glass is 40.0. What is the angle of refraction in water (n w = 1.33; n g = 1.52). 2) If the index of refraction for diamond is 2.42, what will be the angle of refraction in diamond for an angle of incidence in water of 60.0? (n w = 1.33) Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 43 / 57
Law of Refraction The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant (Snell s Law) The incidence ray and refracted ray are on opposite sides of the normal at the point of incidence and all three are in the same plane. When light travels from one transparent medium into another, some reflection always occurs. This is partial reflection and partial refraction. The degrees of reflection and refraction that occur depend on the angle of incidence and the densities of the media. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 44 / 57
Total Internal Reflection and Critical Angle As the angle of incidence increases, the intensity of a reflected ray becomes progressively stronger and the intensity of a refracted ray becomes progressively weaker. As the angle of incidence increases, the angle of refraction increases, eventually reaching a maximum of 90. The point in which refraction ceases and the entire incident light is reflected internally is called total internal reflection. When the angle of refraction is 90, the incident ray forms an angle of incidence that has a unique value for each medium. This unique angle of incidence is called the critical angle. For water, with an index of refraction of 1.33, the critical angle is approximately 49. Watch: Total Internal Reflection What is the critical angle for glass? Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 45 / 57
In general the critical angle can be found using the formula: i c - critical angle n - index of refraction sin i c = 1 n Example: If the index of refraction for zircon is 1.92, what is the critical angle? This is as simple as it gets - a two variable equation! You will have to make use of the inverse sine function once again. sin i c = 1 n = sin i c = 1 1.92 ( ) 1 i c = sin 1 = 31.4 1.92 Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 46 / 57
Practice: Reference your table of indices of refraction where necessary. [See Section 6.3 Video] 1) What is the critical angle in flint glass when light passes from flint glass into air? 2) If the index of refraction for water is 1.33, what is the critical angle? 3) The critical angle for a medium is 40.5. What is the index of refraction of the medium? What is the medium? Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 47 / 57
Section 6.4 - Lenses and Optical Instruments Types of Lenses Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 48 / 57
Refraction of Lenses A lens is a circular piece of glass with a uniformly curved surface that changes the direction of light passing through. When a series of rays passes through a lens, each ray is refracted by a different amount. Rays striking the lens near the edge are bent the most because the curvature is greatest at this point and least at the centre where the surface is flat. In a converging (convex) lens, we have the following parts: Optical Centre (O) - the centre of the lens. Principal Axis (P) - a line drawn through the optical centre perpendicular to the surface. Focus (F) - Parallel rays are refracted and converge at this point. Focal length (f) - The distance from F to O. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 49 / 57
Images Formed by Converging (Convex) Lenses 1) When the object is beyond C (2 focal lengths away), the image produced is between C and F and is real, inverted and smaller than the object. 2) When the object is at C, the image produced at C and is real, inverted and the same size as the object. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 50 / 57
3) When the object is between 1 and 2 focal lengths away, the image is beyond 2 focal lengths and is real, inverted and larger than the object. 4) When the object is at 1 focal length away no image is formed. 5) When the object is between the lens and 1 focal length away, the image is virtual, upright and larger than the object. These 5 cases are identical to the 5 cases for converging mirrors. You will be tested on your knowledge of both, but I will have you only draw situations for mirrors. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 51 / 57
Application: The Human Eye The eye is almost a sphere about 3 cm in diameter. The cornea and lens combine to focus an inverted image on the retina. The cells on the retina respond to various intensities and colours of light and transmit these signals through the optic nerve to the portion of your brain located at the rear of your head. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 52 / 57
Ciliary muscles and suspensory ligaments adjust the shape of the lens to focus images of objects that are far or near on the retina. This ability of the eye to change the focal length of the lens to adjust for far and near objects is called the power of accommodation. The eye can only focus on one image at a time. If it is focused on a near image then images far away will be out of focus. The iris opens and closes the eye opening called the pupil. This change in the pupils opening allows various amounts of light to enter the eye similar to the aperture setting on a camera. The front part of the eye is the cornea. The fluid called vitreous gel maintains the shape of the eye. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 53 / 57
Defects in Vision Farsightedness (hyperopia): is the inability to see nearby objects clearly. It usually occurs because the distance between the lens and the retina is too short or if the cornea-lens is too weak and cannot focus the image on the retina. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 54 / 57
Nearsightedness (myopia): is the inability to see far images clearly. It usually occurs because the distance between the lens and the retina is too long or if the cornea-lens is too strong. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 55 / 57
Correcting Defects in Vision: Watch: Lasik Eye Surgery Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 56 / 57
Astigmatism is the inability of the eye to focus light in different planes at the same time. It is caused when the cornea or lens is not perfectly spherical. Colour Blindness is the inability of the eye to distinguish certain colours. Cataracts are formed due to a gradual loss of transparency of the eye s lens. Mr. Palmarin Chapter 6 - Light, Mirrors, and Lenses 57 / 57