SCP1134 Foundations of Physics - Module 8: Wave Energy (Light) Lecture Notes PDF

SCP1134 Foundations of Physics Module 8: Wave Energy (Light) Module 8: Wave Energy (Light) This module consists of six (6) sections: 1. Introduction and Rectilinear Propagation of Light 2. Reflection 3. Refraction 4. Dispersion a...

SCP1134 Foundations of Physics Module 8: Wave Energy (Light) Module 8: Wave Energy (Light) This module consists of six (6) sections: 1. Introduction and Rectilinear Propagation of Light 2. Reflection 3. Refraction 4. Dispersion and colour 5. Other light phenomena 6. Curved mirrors 7. Lenses. In addition there are separate resources for: a. Answers to exercises and problems b. Readings about Light. Reading Cutnell & Johnson (11th Edition), Chapters 25 to 27. Introduction In Module 6, about Wave Motion and Sound, many principles of wave motions were covered. These principles also apply to this module of Light. Whilst light rays appear to take the shortest route between two places, (actually they take the path of least time), which in any medium (of constant “optical density”) is normally a straight line and is often referred to as “rectilinear propagation of light”, light rays can also be reflected, refracted, dispersed, diffracted, and scattered. Ray diagrams When drawing light ray diagrams there are conventions used (which some text books don’t use because of the printing difficulties). These conventions include: a. A real object and real light rays are drawn with solid lines, while virtual objects or images and virtual rays are drawn with dotted lines. Virtual means that it doesn’t really exist at that place, but can be imagined to exist there because rectilinear propagation of light rays suggests that the light rays should have travelled that pathway. That is, if the light rays were projected on as straight lines, then that is where they would have gone to, or come from. b. If light rays cross and both rays are real rays, a REAL image exists there. If a screen were placed there, the real image would appear on the screen. c. If a light rays cross and one or both are virtual rays, then an image would NOT appear on a screen, however a person looking at a mirror or through a lens will see a VIRTUAL image which only appears to be there, due to rectilinear propagation of light. These conventions will be applied, where possible, in these notes. 1 Edith Cowan University Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) Rectilinear Propagation of Light Street light The ray diagram at right shows how shadow length increases as a person moves further from under a street light. This increase is due to the principle of rectilinear propagation of light. Eclipses can be explained by rectilinear Shadow 1 Shadow 2 Propagation of Light. A Solar Eclipse occurs when the sun is hidden from view from a position on the earth, because the moon is casting a shadow on that part of the earth. The umbra is a region of total shadow, and is often only a few square kilometres in area as it sweeps across the face of the earth. The penumbra is a larger region either side of the umbra, gradually changing from total darkness next to the umbra through to full light. The diagram below, being not to scale, incorrectly shows the penumbra as if its covers the entire earth; it rarely exceeds one-third of the side of the earth facing the sun. Note that the extremities of the umbral and penumbral regions are defined by rectilinear propagation of light. Diagram - Not to Scale Penumbra Moon Sun Umbra Earth A Lunar Eclipse occurs when the moon falls into the shadow region behind the earth as it revolves about the earth about once a month. (Because the plane of the moon’s revolution about the earth is not always aligned with the axis of the earth’s revolution about the sun, a lunar eclipse does NOT occur every month.) Again, the extremities of the umbral and penumbral regions are defined by rectilinear propagation of light. 2 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) Diagram - Not to Scale Penumbra Earth Sun Moon Umbra Section 2 - Reflection of light Reflected ray Incident ray Normal Angle of Angle of incidence (i) Reflection (r) REFLECTION: angle i = angle r Reflecting surface Plane (flat) mirrors will allow an IMAGE to be created, which appears as far perpendicularly behind the mirror, as the OBJECT is in front of the mirror. (This is a consequence of the angle of incidence being equal to the angle of reflection from a plane mirror, no matter at what angle of incidence the rays meet the mirror for reflection) as shown in the diagram below. Object OO Normal Image Example 7.1: Use a diagram to show that a person can see all of him/herself in a correctly positioned vertical mirror, which is exactly half the person’s height. Answer 7.1: Step 1 - Draw in the person (solid lines because the person is real), and the mirror (with hachuring on the back of the mirror, representing the “back-silvering”). Then draw in the image, so that each part of the image 3 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) is to be as far behind the mirror perpendicularly as the object is in front of the mirror. As the image is virtual it should be dotted. Mirror height needed is half the person’s height. Step 2 - Put in an enlarged eye on the object’s head. Step 3 - Draw a line (dotted behind the mirror, solid in front of the mirror) from the image’s foot to the top of the eye. Repeat for the image’s foot to the bottom of the eye. Step 4 - Where the step 3 rays met the mirror, join with solid lines to the object’s foot, making sure the arrows go from the foot towards the mirror (the way the real light rays would actually travel). Step 5 - Repeat steps 3 and 4 for the rays from the image’s top of the head to the top and bottom of the eye, then solid lines from the object’s top of the head to the mirror, again with arrows on the real rays. [The diagram is finished, by adding any labelling that is required.] Because the image is as far behind the mirror as the object is in front, the mirror is half the distance between object and image, so a mirror half the object’s height would be satisfactory for a person to see an image of themselves, if the mirror was correctly positioned. NB 1 - At each reflection at the mirror’s surface in the above diagram, the angle of incidence equals the angle of reflection at that position, however to label that as such would clutter the diagram. NB 2: In the above example, an arrow is placed on the real light rays showing the direction of propagation from the source of light (the object) towards a real image or towards the viewer’s eye (if a virtual image). Important Comment  Why are 2 rays needed, one to the top of the eye and one to the bottom of the eye in the previous answer? Some textbook authors try to get away with a single ray, but any of you who have taken camera photographs, which haven’t been correctly focussed, will know that a blurred image results. It is from the slightly diverging light rays that enter the eye’s pupil (or the camera’s lens), which are then converged and focussed onto the eye’s retina (or the camera’s film), that causes the rays to be “focussed”. The strain in the person’s ciliary muscles (or the camera’s distance setting) allows distance to be judged. If an object is distant, the light rays entering the eye are almost (but not quite) parallel, while for a very close object, the light rays travelling towards the eye are more divergent, so more focussing power is needed to get the focussed image on the retina. Drawing the pair of light rays is often called drawing “focussed rays”, or “a focussed ray diagram”. Lateral Inversion When looking into a mirror to comb your hair, if you are right-handed, the image looking at you appears to be using its left hand. This is called LATERAL INVERSION. 4 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) Example: Draw a diagram to illustrate the lateral inversion of themselves, that a person sees when combing his/her own hair. Comb is in the person’s Comb is in the Image’s RIGHT hand LEFT hand. The diagram above shows the person (on the left) facing the mirror has the comb in the person’s right hand, while the image, (every part of which is the same perpendicular distance behind the mirror as the object is in front of the mirror), is also facing the mirror, and has the comb in the image’s left hand - so illustrating LATERAL INVERSION. [Note that drawing the light rays is NOT necessary to obtain a satisfactory answer in this question.] Section 3 - Refraction of light Refraction occurs because the velocity of the wave motion changes. Rays which are NOT normal (i.e. NOT at right angles) to the surface between the two media will bend as the velocity of the wave motion changes. A change in wavelength occurs as a result of the refraction. When light rays enter a denser medium (e.g. from air to water), the refracted ray is refracted towards the normal in the denser medium, and on the opposite side of the normal from the incident ray, as shown by “ray 1” in the diagram below. (The normal is a line which is perpendicular to the surface separating the two media) Normal Incident ray 2 Incident ray 1 λ1 Less dense medium Angle of incidence Wavefronts are closer together in Angle of deviation the denser Angle of λ2 medium, i.e. refraction wavelength Denser decreases. medium Refracted ray 2 Refracted ray 1 Rays which are incident normally (that is, perpendicular to the surface) as with “ray 2” above, do NOT show bending, but because the velocity in the denser medium is less that the velocity in the less dense medium, the wavelength in the denser medium is reduced, so 5 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) the ray has still been refracted. Wave fronts show that refraction has occurred because the wavelength has decreased in the denser optical medium. The diagram above shows: 1. the angle of incidence as being between the incident ray and the normal, 2. the angle of refraction as being between the refracted ray and the normal, 3. the angle of deviation being the angle through which the incident ray has been deviated from where it would have travelled (shown dotted) and the refracted ray, and 4. for ray 2, both the incident and refracted rays are on the normal, so there is no angle of deviation. 5. for ray 2, the wavelength in the denser medium (λ2) is less than in the less dense medium (λ1). Refractive Index The refractive index of a medium can be defined in two equivalent ways: i. The refractive index of a medium is given by n 1.sinα 1  n 2.sinα 2  n 3.sinα 3 ... , where n 1 , n 2 , n 3 , etc are the refractive indices and α1 , α 2 , α 3 , etc are the angles between the normals and the rays for the media 1, 2, 3, etc.. ii. The refractive index of a medium is given by the ratio of the velocity of light in vacuo (i.e. in a vacuum) compared to the velocity of light in that medium. c1 i.e. 1n2  c2 where 1 n2 is the refractive index for light from medium 1 passing to medium 2, c1 is the velocity of light in medium 1 and c 2 is the velocity of light in medium 2. NB1: The velocity of light in vacuo is often denoted as c 0. NB2: The velocity of light in the earth’s atmosphere is very close to c 0 , and only in the most accurate work does the difference need to be considered. As a consequence, the refractive index from vacuo to air ( 0 n air ) is usually taken as being 1.00 Example 7.2: If the refractive index of water (relative to air) is 1.333, a. what is the angle between the normal and an refracted ray into the water, for a ray of incident angle of 60°, and b. through what angle has the light ray been deviated, and c. what is the velocity of light in water? (Use velocity of light in air as 3.00  10 8 m s 1 ). Answer 7.2: (a) n 1.sin α1  n 2.sin α 2 |Air: 0 n 1  n 1  1.000 and α1  60 1.000  sin 60  1.333  sin α 2 |Water: 0 n 2  n 2  1.333 and α 2  ? 1.000  sin 60 0.866  sinα 2    0.650 1.333 1.333  α 2  40.5 The angle between the refracted ray and the normal in water is 40.5° 6 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) (b) The angle of deviation = angle of incidence - angle of refraction. = 60° - 40.5° Angle of deviation is 19.5° c0 (c) 0 n1  | 0 n1  n 1  1.333 c1 3.00  10 8 1.333  | c 0  3.00  10 m s 8 1 c1 3.00  10 8 c1   2.25  10 8 m s 1 | c1  ? m s 1 1.333 Velocity of light in water is 2.25 ×108 m.s-1 Labelled diagrams illustrating refraction can be drawn Example 7.3: Show that a vertical stick appears bent at the water’s surface, when an observer is standing to one side. Answer 7.3: Vertical stick Air Water Note how the section of stick underwater appears bent towards the observer. To draw this diagram and get it looking reasonably correct, the steps are: 1. Draw the water surface, the eye and the vertical stick. Label stick, water and air. 2. Locate the bottom of the stick’s image about three-quarters depth and slightly on the eye’s side from where the real bottom of the stick is located. 3. Draw solid arrowed rays from the top of the stick to the top and bottom of the eye. 4. Repeat step 3 for the water surface level of the stick. 5. Draw the 2 parts, dotted rays in the water and solid arrowed rays in air, for the one straight line from the bottom of the image to the top of the eye. 6. Draw a solid arrowed ray from the bottom of the real object to the water surface where the rays in step 5 crossed the water surface. 7. Repeat steps (5) and (6) for the rays to the bottom of the eye. 7 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) Light Refraction can occur gradually (as Imagined direction from sun when light rays enter the earth’s atmosphere at an oblique angle) with the rising or setting sun. In the diagram at right, we have to imagine Sun’s Rays Earth that the atmosphere has spherical layers around the earth, (two layers are shown) and Diagram is as a light ray gets closer to the earth it meets NOT to scale these denser air layers and is refracted slightly downwards (towards the “normal”) and a person on the horizon believes they are seeing the setting (or rising) sun in the direction labelled as “Imagined direction from sun”, but really the sun is lower and probably below the person’s true horizon. For explanatory purposes the diagram is exaggerated, but in reality, the total bending of the sun’s rays is only about one degree, corresponding to an observed sunrise about 4 minutes before true sunrise, and observed sunset about 4 minutes after true sunset. In reality, the bending of the sun’s rays is a gradual process as the rays enter progressively denser air closer to the earth’s surface. Critical Angle Suppose a torch were held under water as a light source and a narrow beam, such as a ray were pointed towards the surface. When pointed vertically upwards, the ray would emerge into the air without deviation. If the ray met the water surface obliquely, the ray in air would be deviated away from the normal. As the ray met the water surface more obliquely, a point would be reached where the emergent ray into air was at “grazing angle”, that is, parallel to the water surface. Beyond this critical angle in the water, total internal reflection will occur in the water, as shown in the diagram below. Ray normal to surface Rays deviated away emerges without from their normals on deviation, but is still emerging into less refracted. dense medium. Normals Grazing ray Totally internally Torch reflected ray angle i = angle r Critical angle Normal for Total Internal Reflection Partial internal reflection occurs even for rays that emerge into the air, (but this is rarely indicated in diagrams, because it tends to clutter the diagram), but when the critical angle is exceeded, the internal reflection is total. 8 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) Example 7.4: Given that the refractive index from air to water is 1.333, what is the critical angle in water? Answer 7.4: n 1.sin α1  n 2 sin α 2 |For water 0 n 2  1.333 and α 2  ? 1.000  sin 90  1.333  sin α 2 |For air 0 n 1  1.000 and α1  90 1.000  sin 90  sin α 2   0.750 1.333  α 2  48.6 The critical angle for water is 48.6° Section 4 - Dispersion and Colour Dispersion is the term used to describe how wave motions of different frequencies are separated into different frequencies as they are refracted by slightly different amounts. Dispersion of light is the cause of coloured fringes on the edges of a projection screen if the projector is out of focus, and of the colours seen in cut glass, diamonds, crystal glasses and vases, and in some transparent plastics. Dispersion will also occur with lenses causing “chromatic aberration”, resulting in each colour of the spectrum having slightly different focal length, so a projected coloured image is often difficult to focus. Diagram NOT Angle of deviation to scale White light R O Y G B Angle of I dispersion V Dispersion of white light through a glass prism The above diagram is not to scale, because in reality, the angle of dispersion is only about 1° or 2°, while the angle of deviation is around 40° to 50° for a glass prism with a vertex of 60°. Different types of glasses (e.g., crown glass, flint glass, crystal and plastics have slightly different refractive indices, so their deviating ability and dispersing ability differ). Beyond the visible red end of the spectrum is infra-red (“I/R”) radiation (heat rays), while beyond violet is invisible ultra-violet (“U/V”), supposedly responsible for some types of skin cancers on humans. Newton showed that sunlight could be separated into the spectrum by use of a prism, and that a second prism would produce no more colours, although the angle of dispersion would increase. 9 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) Red White White Violet White light Wide slit opening Newton also showed that the colours could be reformed to produce white light, as shown above, when light passing through a wide slit opening has passed through two prisms, one with its vertex at the top and the other with its vertex at the bottom, will reform white light in the middle, but with a red coloured edge one side, and a violet coloured edge the other side. To clearly see a full colour spectrum on a screen from dispersion, it will be necessary to have a well darkened room, and only a narrow beam of white light entering the prism, (such as through a narrow slit, so the colours from different initial white rays do not recombine to produce a white image with coloured fringes). Colour Subtraction Colour subtraction occurs when particular frequencies (colours) of light are absorbed by pigments, and the remainder are reflected (if the pigmented object is an opaque reflector), or transmitted (if the pigmented object is transparent, and then it is called a “colour filter”). That is, the pigment subtracts (removes) colours and allows the colour that is reflected or transmitted to be the colour seen. Colour subtraction occurs when coloured objects are illuminated by white or coloured lights, and to mixing of coloured paints. Because pigments in nature are rarely “pure”, they not only transmit or reflect their own colour, but they usually also transmit (if a filter) or reflect (if opaque) the spectral colour either side of them, depending on the purity and density of the pigment. Example 7.5: White light produces a spectrum of (I/R) R O Y G B I V (U/V) A red rose reflects (I/R) R O and absorbs Y G B I V (U/V) [Try these for yourself - answers are in the “Answers” file.] A green leaf reflects and absorbs A blue filter transmits and absorbs A black object reflects and absorbs Example 7.6: When coloured light falls on a coloured object, the resulting colour of the object may change. For example, if a yellow cloth is illuminated by blue light, The yellow cloth would normally reflect O Y G and absorb other colours 10 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) But the blue light probably contains only G B I and the B+I would be absorbed by the yellow cloth so the only colour reflected is G Therefore the yellow cloth now appears Green Example 7.7: A painter mixes red and yellow paints, what colour might be expected to result? [Try this yourself - answer in the “Answer” file] The red paint will reflect The yellow paint will reflect The mixture of red and yellow paints will reflect and so this mixture of paints will appear. Colour Addition Colour addition occurs when different colours of light fall on an object. This can occur in stage drama performances when different coloured projectors are used. There are three “Primary colours”, “Primary Red”, “Primary Green” and “Primary Blue”, which are of particular frequencies. If equal intensities of these three primary colours are used to illuminate a white screen, then white light will be produced. The “Colour Triangle” is a model that helps understand colour by RED addition; (1) R + G + B = White Secondary colours are formed by yellow magenta addition of two primary colours. (2) R + G = Yellow WHITE (3) R + B = Magenta (4) G + B = Cyan GREEN cyan BLUE The “complement” of any primary or secondary colour is the colour opposite it in the colour triangle, so: the complement of RED is CYAN and the complement of CYAN is RED the complement of GREEN is MAGENTA and the complement of MAGENTA is GREEN the complement of BLUE is YELLOW and the complement of YELLOW is BLUE and equal intensities of any colour with its complement will produce WHITE. Section 5 - Other light phenomena (1) Scattering Scattering may be regarded as the irregular reflection of a wave motion due to the presence of millions of minute reflective particles. The extent of scattering depends upon 11 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) the wavelength of the wave motion compared to the size of the particles doing the scattering. Light can be scattered by dust particles, R minute ice crystals and water droplets in the R B air. Because blue light is of higher frequency than red light, blue light has shorter wavelength and blue tends to be scattered more than red light. Earth At sunrise or sunset, sunlight passes obliquely through a thick layer of the earth’s atmosphere, where the blue light is scattered more than the red, so looking towards the sun the light seen is deficient in the blue end of the spectrum and the rising or setting sun appears orange- red, but looking vertically above, an observer sees light rays that have passed far overhead and more of the scattered blue light is seen, so the overhead sky appears blue, but the rising or setting sun appears reddish. (2) Diffraction and Interference (Not examinable for SCP1134) Diffraction does occur with light, as evidenced by a diffraction grating, which can be used to produce a colour spectrum. A diffraction grating consists of many very finely spaced parallel lines, each of which produces a diffraction pattern, and interference between the diffraction patterns from the many finely spaced parallel lines will produce a spectral pattern. The superposition (which is a result of interference) of one frequency wave on another in different ways is used in AM (amplitude modulation) and FM (Frequency modulation) radio transmissions, which have been covered in the Readings for Module 7. (3) Polarisation of Light (Non-examinable for SCP1134) Suppose light is being reflected obliquely off a surface, (for instance about an hour before or after sunset looking over still ocean water), the transverse light wave components of the oscillations which are horizontal are reflected with little change in amplitude, but the vertical oscillation components have greater absorption by the reflecting surface. This can lead to observers suffering from “glare”. Polaroid spectacles have polarizing filters which reduce the intensity of the horizontal light oscillations, hopefully bringing them closer to the amplitude of the vertical light wave oscillations and making the observer perceive less glare. Light reflected from a very smooth road can be polarised and cause discomfort to drivers due to the glare. Wearing “polaroid” glasses can reduce this problem of glare. (4) The “Dual Nature of Light” (This IS examinable for SCP1134) Rectilinear propagation of light, reflection, refraction, dispersion, diffraction, interference, scattering and polarization of light can all be satisfactorily explained by the WAVE THEORY of light. However, the wave theory can NOT satisfactorily explain the photo- electric effect of light. In the photo-electric effect, electromagnetic radiation of a particular frequency striking certain electrical conductors or semi-conductors will cause ejection of electrons from that material, and that flow of electrons is an electric current. To explain how light energy is 12 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) converted to electricity by this photo-electric effect, the PHOTON theory of light has been developed, in which particles, known as “photons”, “corpuscles”, or “particles” or “bundles of energy” are the basis of an electro-magnetic radiation. There is a threshold frequency that the radiation must exceed before any electron flow occurs, and this threshold frequency is different for different materials. It is now known that the energy of a wave motion is related to its frequency by the formula E  h.f , where “h” is Planck’s constant. However, this photon theory does NOT satisfactorily explain some of the phenomena which can be explained by the Wave Theory. Neither the Wave Theory nor the Photon Theory, on its own can give a complete and satisfactory explanation of all the observed phenomena of light. The Quantum Mechanics theory uses a mathematical equation known as the Schrödinger Wave Mechanics equation to explain electro-magnetic propagation. It incorporates both wave and photon explanations of light phenomena. As this equation involves second order differentiation in calculus, it is beyond the scope of SCP1134. The SCP1134 course concentrates on those aspects of light which can be explained by the wave theory of light, but acknowledges that the photo-electric effect (which is used in many electronic sensors) does exist. The “DUAL NATURE” of light acknowledges that light has properties of both waves and particles and this is currently explained by Quantum Mechanics. Section 6 - CURVED MIRRORS Curved mirrors usually have their reflecting surface as part of a sphere (hence “spherical mirrors”), or part of an ellipsoid (“elliptical mirrors”). Elliptical mirrors give clearer focussing of light rays, so are often used in vehicle headlights and in certain types of telescopes. Curved mirrors can be CONVEX (reflecting surface bulges outwards, as when looking at the underside of soup spoon) or may be CONCAVE (reflecting surface is “hollowed”, as when looking into the top of a soup spoon). Curved Mirror Terminology (applied to a CONCAVE mirror). Light ray parallel Optical centre of to principal axis mirror Principal axis r C aperture Rear surface of the concave mirror is “hachured”, by dotted lines F f Optical centre of a mirror is the middle of the reflecting surface. Centre of curvature (C) of a spherical mirror is the centre of the sphere of which the mirror surface is part of that sphere. 13 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) Radius of curvature (“r”) of a mirror is the distance from the centre of curvature to the mirror surface. Principal axis of a mirror is a line that joins the mirror’s optical centre and centre of curvature and can be extended in either direction. Aperture of a mirror is the width of the reflecting surface. Principal Focus (F) of a mirror is the point on its principal axis, where all light rays that are parallel to the principal axis before reflection, will pass through after reflection (concave mirror), or appear to diverge from (convex mirror). Focal Length (“ f ”) is the distance from the Principal Focus to the Optical Centre of the mirror. r NB: For a spherical mirror of small aperture, f  2 Curved Mirror Light Ray Diagrams 1. Light rays parallel to the principal axis, a. for a concave mirror, will pass through the principal focus in front of the mirror after reflection. b. for a convex mirror, appear to diverge from the principal focus behind the mirror after reflection. 2. Light rays passing through the centre of curvature of concave mirror (and those directed towards the centre of curvature of a convex mirror) are reflected back on themselves. 3. Light rays passing through the principal focus of a concave mirror are parallel to the principal axis after reflection. Mirror Plane O Ray 1 F Ray 3 I Ray 2 C Example: The above three rules are illustrated below for an object 2.0 cm high, which is 10.0 cm from a concave mirror of focal length 4.0 cm. NB 1: Reflection is drawn to occur at a “mirror plane” rather than the curved surface of the drawn mirror. This gives better accuracy in the image location, because the curvature of mirrors is usually exaggerated in diagrams. 14 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) NB 2: The image is located where the three rays are concurrent. Any two of the three rules should lead to the location of the image. NB 3: In this case, a real image was formed, because the light rays actually travelled the paths shown. If it was necessary to extend one or more rays behind the mirror to get concurrency, (because the rays can’t pass through the solid mirror), they would be dotted “virtual” rays and the image would be “virtual”. Virtual images are common with convex mirrors. NB 4: Rays from the top of the object locate the top of the image. Because the object’s base stands on the principal axis, the image’s base must also be on the principal axis (all three ray rules would require this). It could be shown that rays from half-way up the vertical object would be half-way between the image’s top and the principal axis, but this would clutter the diagram. To adequately describe the image formed in the example above a number of statements need to be made: (1) Is the image real or virtual? In the example above it is real (2) Is the image upright or inverted? In the example above it is inverted (3) Is the image magnified, same size or diminished? It is diminished (4) Where is the image located? The image is 6.7 cm in front of the mirror (5) How high is the image? The image is 1.3 cm high Stating these five pieces of information in a single sentence: “The image is real, inverted and diminished, 6.7 cm in front of the mirror and 1.3 cm high”. Calculation Example: The image type and position can be obtained by accurately drawn scale diagrams, however 1 1 1 the formula   can be used to calculate the image distance and type (real or p q f virtual), where p = object distance, q = image distance, and f = focal length, and using a sign convention that “real distances” are positive, and “virtual distances” are negative. NB: The focal length is positive for converging lenses (convex lens) and converging mirrors (concave mirror). The focal length is negative for diverging lenses (concave lens) and diverging mirrors (convex mirror). q The formula: Magnification    can be used to find if the image is magnified, same p size, or diminished, and whether the image is inverted or upright. If the magnification is positive, then the image is upright, if the magnification is negative, then the image is inverted. 1 1 1 For the Example on pages 16,17   | p  10.0 cm p q f 1 1 1   | q  ? cm 10.0 q 4.0 1 1 1 52 3      | f  4.0 cm q 4 10 20 20  q  203  6.67 cm 15 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) and because the answer is positive, the image is real. q  20 1  2 Magnification           p  3 10  3 This negative sign means the image is inverted, and the image height is 23 the height of the object, therefore the image is diminished and its height is 23  2.0 cm  1.33 cm The image is real, inverted and diminished, 6.7 cm from the mirror and 1.3 cm high. What do these images look like? 1. Look into the bowl of a shiny soup spoon which you hold 10 to 15 cm in front of your face and observe the image you see. Because the curvature on most spoons is different in the horizontal and vertical planes as you hold the spoon, the image will be a little distorted. You can’t see whether it is real or virtual, but you can see it is inverted and diminished, although the image’s location may be hard to judge. This is the kind of situation covered in the example above. 2. Now bring the spoon very close to your face, so the spoon almost touches your eyebrow. You should be able to see a magnified image, which by moving the spoon around a bit, has your eyebrows above your eye, so the image is now upright. The next example treats this kind of image. Example Describe the image formed when an object 1.0 cm high, is 2.0 cm from a concave mirror of focal length 4.0 cm. Notice that when the image distance is less than the focal length, the rays passing through F and C will never meet on the same side as the object, so the image must be behind the mirror, which means the light rays never actually travel there, just their “propagation” extensions, so the image is virtual. The image (by measurement and observation) is virtual, upright and magnified, 4.0 cm behind the mirror and 2.0 cm high. Mirror plane I O C F Calculation can be used to verify the accuracy. 1 1 1 In this Example   | p  2.0 cm p q f 1 1 1   | q  ? cm 2.0 q 4.0 16 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) 1 1 1 1 2 1      | f  4.0 cm q 4 2 4 4  q  14  4.0 cm and because the answer is negative, the image is virtual. q   4.0 1  2 Magnification          , that is magnified, p  1 2.0  1 and being positive, the image is upright The image height is twice the height of the image, i.e. 2  1.0 cm  2.0 cm The image is virtual, upright magnified 4.0 cm from the mirror and 2.0 cm high. The previous example is used in a dentists’ mirrors. Dentists need an upright and magnified image of the person’s tooth. What do these images look like? 3. One further situation can be tried while you have the soup spoon. Hold the spoon and turn it over so you are looking at the bulging underside of the spoon. An inverted and diminished image should be observed. You will come across this situation in one of the tutorial problems. The uses of curved mirrors are discussed in the Readings for Light. Section 7 - Lenses Lenses produce images by refraction of light rays. (The singular of lenses is “lens”). Lenses are described by their shape and by their function (i.e. converging or diverging.) Convex Lenses are thicker at their centre than at their edges, and they will cause light rays to converge. Double convex Plano convex Meniscus convex Bi-convex Concavo-convex Concave Lenses are thicker at the edges than at their centre, and they will cause light rays to diverge. Double concave Plano concave Meniscus concave Bi-concave Convexo-concave 17 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) NB: For spectacles, the curved face of a lens could be shaped as part of a sphere (in cheaper lenses), as part of an ellipsoid (more expensive lenses), or some modification of a sphere or ellipsoid to match the person’s optical need. Lens Terminology Optical centre of a lens is the point through which a light ray passes without deviation. For bi-convex and bi-concave lenses, this is the middle of the lens. Principal axis of a lens is a line that passes through its optical centre and is perpendicular to the lens. The principal axis is extended in either direction. Aperture of a lens is the width of the lens. Principal Focus of a lens is the point on its Principal Axis through which all light rays that are parallel to the Principal Axis before refraction, will pass after refraction (if a convex lens), or appear to diverge from (if a concave lens). NB: A lens has Principal Foci on both sides, and the two focal lengths are equal in magnitude. Focal Plane is an imaginary line at right angles to the Principal Axis, passing through the Principal Focus, on which parallel light rays (that are NOT parallel to Principal Axis) will pass after refraction from a convex lens, or diverge if a concave lens. Focal Length (“f”) is the distance from a Principal Focus to the Optical Centre of the lens. Rules for Lens Light rays 1. Light rays parallel to the Principal Axis pass through the Principal Focus on the other side of the lens from the object after refraction through a convex lens, or appear to diverge from the Principal Focus on the same side of the lens as the object after refraction through a concave lens. 2. Light rays passing through the Optical Centre are not deviated. 3. Light rays passing through the Principal Focus of a convex lens travel parallel to the Principal Axis after refraction. 18 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) Example: A student observes a 1.0 cm long insect through a convex lens. If the insect is 3.0 cm from the lens and the lens has a focal length of 5.0 cm, where would be the image of the insect? How many times larger than the real insect would be the image? Answer (The answer could be either a calculation OR a scale diagram. Both solutions are given below). Diagram solution: NB. Not all three rays were required to obtain the image position. In many diagrams any two of the ray rules will produce the image position. The image is virtual, upright and magnified, 7.5 cm from the lens the same side as the insect, and it is 2.5 cm high. 1 1 1 (2) Calculation   | p  3.0 cm p q f 1 1 1   | q  ? cm 3.0 q 5.0 1 1 1 35 2      | f  5.0 cm q 5 3 15 15  q  215  7.5 cm and because the answer is negative, the image is virtual. q   15.0 1   5 5 Magnification              2.5 p  2 3.0   2  2 Because Magnification = +2.5, the image is upright and height is 2.5 times the 1.0 cm height of the insect, so the image is magnified, and the image height is 2.5 cm. The image is virtual, magnified and upright, at a distance of 7.5 cm from the mirror and is 2.5 cm high. With a different location of the object relative to the focal length, upright and inverted images, real and virtual images can be obtained with lenses and mirrors. Some situations have been encountered in these notes, while others will be encountered in the tutorial problems, or can be found in the “Readings for Light”. 19 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) Module 8 - Light - Questions and Problems Section 1 - Introduction and Reflection Question 1: Explain why light from a torch only carries a certain distance in front of you as you walk. Question 2: When looking across a still lake, reflections of buildings the other side of the lake can be seen in the water: a) are the reflected images real? b) are the reflected images smaller or larger than the real buildings? c) are the reflected images upright or inverted? d) Draw a diagram to illustrate how the image is seen by an observer from the other side of the lake. Problem 1: The dial of one radio is calibrated by frequencies and on another radio dial by wavelength. Both radios are tuned to the same radio station. If one dial reads 94.1MHz, what should the other one read? Problem 2: Radio station 1080 AM (used be 6IX) transmits on a frequency of 1080 kHz. -1 Given that the velocity of electromagnetic radiation is 3.00 x 108 m s , calculate the wavelength and period of the transmitted signal. Section 2 - Refraction Question 1: With aid of a diagram explain why a person standing at the side of a swimming pool sees the pool’s bottom to be shallower than it really is. Question 2: Draw two wave front diagrams, to show refraction when wave fronts are parallel to the surface and when wave fronts are oblique to the surface being encountered. Question 3: A scuba diver looks up to the water surface whilst swimming under water. Describe what the diver would see as (s)he looks vertically upwards and then down horizontally in any direction. Section 3 - Dispersion and Colour Question 1: Make simple copies of the following diagram. On each mark how it would appear to an observer if illuminated: a) by a red light. b) by a blue light red c) by a green light. d) through a cyan filter. Blue e) through a blue filter. Green Yellow f) through a magenta filter. Question 2: When viewing a cut glass vase in sunlight, the glass often appears to sparkle with coloured lights. Explain why this occurs. Question 3: If you have your back to the sun when looking towards dark clouds, you can sometimes see a rainbow. Explain why this can occur. Question 4: Complete and label the following diagrams to show the difference between simple refraction and dispersion. 20 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) Simple Refraction Dispersion Section 4 - Other light phenomena Question 1: There are two current scientific models for light. a) What are they? b) What are the strengths of each? c) What are the weaknesses of each? Question 2: What are some of the special properties of: a) U/V light? b) X Rays? c) Cosmic rays? Question 3: Why is the setting sun near the horizon red, while the sky overhead is blue? Section 5 - Curved mirrors Problem 1: Describe the image formed when an object 2.0 cm high, is 5.0 cm in front of a convex mirror of focal length 3.0 cm. Section 6 - Lenses Question 1: A student holds a convex lens near the laboratory wall and moves it back and forth until a clear image of the window is seen on the opposite wall. The lens is then 10 cm from the wall. a) Is the focal length of the lens approximately 5,10, 20 or 40 cm? b) Strictly, is the focal length slightly greater than or less than your answer to (a)? Explain. Question 2: Which of the following statements are correct about the image formed by a concave lens? a) It is real. b) It is virtual. c) It is closer to the lens than the object. d) It is further from the lens than the object. e) The magnification is greater than one. f) The magnification is less than one. g) It is formed on the opposite side of the lens from the object. h) It is formed on the same side of the lens as the object. i) It is upright (relative to the object). j) It is inverted (relative to the object). 21 Copyright © Dr. S. Hinckley, ECU SCP1134 Foundations of Physics Module 8: Wave Energy (Light) Problem 1: Describe the image formed when a convex lens of focal length 3.0 cm, has an object 2.0 cm high is placed 5.0 cm in front of the lens. Problem 2: To photograph the detail on a plant leaf a student uses extension tubes to increase the distance between the camera lens and the film at the back of the camera. The convex lens has a focal length of 30mm and is 60mm from the leaf in order to produce a focussed image on the film. a) How far away from the film is the lens? b) What is the magnification of the image? Problem 3: Describe the image formed by a concave lens of focal length 6 cm. 22 Copyright © Dr. S. Hinckley, ECU