Physics Optics 2 Refraction, Summaries of Physics

Download Physics Optics 2 Refraction and more Summaries Physics in PDF only on Docsity! Chapter 4: Refraction Please remember to photocopy 4 pages onto one sheet by going A3-. A4 and using back to back on the photocopier Something to think about “Twinkle twinkle little star...” Why do stars twinkle but not planets? Refraction is the bending of a wave as it passes from a medium of one refractive index to another medium of different refractive index. When light travels from a medium of lower refractive index to a medium of higher refractive index it is refracted towards the normal and vice versa*. Emergent Ray Air Glass Normal ---- aa | Incident Ray The two laws of refraction of light 1. The incident ray, the normal and the refracted ray all lie on the same plane. (not to be confused with the reflected ray, from the laws of reflection of light). 2. The ratio of the sin of the angle of incidence to the sin of the angle of refraction is a constant called the refractive index. Ry .=ot sinr The refractive index between any two media is a constant and is given the symbol y. The second law of refraction is also known as Snell’s law*. This leads to the following definition: The refractive index of a medium is the ratio of the sine of the angle of incidence to the sine of the angle of refraction when light travels from a vacuum into that medium. (In practice we consider air and a vacuum to be equivalent). Note that if you see the phrase “the refractive index of glass is 1.5”, it means that “when light travels from air into glass the refractive Index is 1.5”. This is written as alg = 1.5 If light is going in the other direction (i.e. from glass to air), the refractive index will be the inverse of 1.5. 1 15 Le. if ang = 1.5, then nm, = 1 If ang = 1.5, then na = 75 See 2012 Question 12 (b) {I think this was the only time it arose, which was very nasty because most people wouldn’t be expecting it} Applications of Total Internal Reflection 1. Optical Fibres 2. Endoscopes 3. Reflective road signs 4. Periscopes Optical Fibres* Cladding (72) Core (V1) yé= ql How light is transmitted in an optical fibre 1. An optical fibre consists of a glass pipe coated with a second material of lower refractive index. 2. Light enters one end of the fibre and strikes the boundary between the two materials at an angle greater than the critical angle, resulting in total internal reflection at the interface. 3. This reflected light now strikes the interface on the opposite wall and gets totally reflected again. 4. This process continues all along the glass pipe until the light emerges at the far end. Note: e If the second cladding material wasn’t there or had a refractive index greater than that of the central core total internal reflection would not occur and the light would simply escape out. e — The outer cladding also acts as a protective layer against scratches etc. e The word "fiber" comes from the fact that these cables are incredibly thin, about the size of a hair. Applications Telecommunications Medicine (endoscopes) Advantages of optical fibres over copper conductors Less interference / cheaper raw material / occupy less space / more information (carried) in the same space / flexible for inaccessible places/ do not corrode R.1.= Summary of Formula R.r.=sinl R.L= Speed of light€ air R.L= 1 sinr S peed of light€ medium sinC —_ Real Depth Apparent Depth 1 If ang = 1.5, then Na = Ts Mandatory Experiment: To verify Snell’s Law and use it to measure the refractive index of a solid. Leaving Cert Physics Syllabus: Refraction Content Depth of Treatment Activities sTSs 1. Laws of Refractive index Demonstration using ray box Practical examples, e.g. real and refraction or laser or other suitable apparent depth of fish in water. Refractive index in terms of relative speeds. method. Appropriate calculations. Appropriate calculations. 2. Total internal Critical angle Demonstration. Reflective road signs. reflection Relationship between critical | Appropriate calculations Mirages. angle and refractive index. Prism reflectors. Transmission of light Uses of optical fibres: through optical fibres. ¢ — telecommunications e medicine (endoscopes) Experiments: 1. Verification of Snell’s law of refraction. 2. Measurement of the refractive index of a liquid or a solid. VERIFICATION OF SNELL’S LAW OF REFRACTION Or TO MEASURE THE REFRACTIVE INDEX OF A GLASS BLOCK APPARATUS: Glass block, ray-box, protractor, page Emergent Ray DIAGRAM (I couldn’t find one which included the ray-box) Air Glass Air PROCEDURE 1. Place a glass block on the page and mark its outline. Normal ~~ 2. Shine a ray of light from the ray-box into the glass block. 3. Mark two dots on the incident ray and exit ray and draw the outline of the block. Incident Ray 4. Remove the block and complete all lines including the normal, as indicated on the diagram. 5. Measure the angle of incidence i and angle of refraction r using the protractor. 6. Repeat for different values of i. 7. Draw upa table as shown. 8. Plot a graph of sin i against sin r. A straight line through the origin verifies Snell’s law of refraction ie. sini « sinr. 9. The slope of the line gives a value for the refractive index of glass. 10. The refractive index of glass is also equal to the average value of ant RESULTS i (degrees 1 (degrees sini sinr sini n= sinr Average value for refractive index = SOURCES OF ERROR / PRECAUTIONS 1. Using small angles of incidence will result in large percentage errors. 2. Place two dots far apart on the incident and refracted light beams to accurately locate the beams Tip when carrying out the experiment Start with a small angle of incidence and measure the corresponding angle of refraction. Now increase this angle at roughly regular intervals each time to ensure a wide range of values (rather than simply taking angles at random each time). Experiments There are three mandatory experiments on the syllabus to do with refraction To verify Snell’s Law To measure the refractive index of a liquid or a solid. The or in italics implies that you must be given the option, and therefore the exam question cannot specify measuring the refractive index of a liquid. But in verifying Snell’s Law, we plot a graph of Sin i against Sin r, and show that because the graph is a straight line going through the origin, the two variables are directly proportional. To calculate the refractive index we then simply calculate the slope of the graph, and if Sin i is on the y-axis, the slope corresponds to the refractive index. No need to worry about measuring the refractive index of any messy liquids! 10 4. 5. 6. @ Exam questions: Refraction [2010 OL] Which of these scientists is associated with the law of refraction of light? Rutherford Snell Joule Einstein [2008][2006][2002 OL][2004 OL][2005 OL ][2009 OL] What is meant by refraction of light? [2008]State Snell’s law of refraction. air | cornea [2002 ][2002 OL]State the laws of refraction of light. [2004 OL]What is meant by the refractive index of a material? normal _ > [2008] Light is refracted as it travels from air into the cornea as shown in the diagram. Calculate the refractive index of the cornea. (ii) Draw a diagram to show the path of a ray of light if it travelled from water of refractive index 1.33 into the cornea. 7. @ (ii) In the diagram the value of the angle i is 41.8°. Calculate a value for the refractive [2005 OL] What special name is given to the angle of incidence i, when the effect shown in the diagram occurs? index of the glass. (iii) Draw a diagram to show what happens to the ray of light when the angle of GLASS 8. 9. 10. 11. 12. 13. 14, 15. @ incidence i is increased to 45°. [2003 OL] Explain total internal reflection with the aid of a labelled diagram [2004 OL] When will total internal reflection occur? [2004 OL] Define the critical angle. [2003 OL] The critical angle for the glass is 42°. Calculate the refractive index of the glass. [2007] The refractive index of a liquid is 1.35, what is the critical angle of the liquid? [2003] Calculate the critical angle for diamond. The refractive index of diamond is 2.4. [2010] What is the critical angle of a sample of glass whose refractive index is 1.46? [2003 OL] The diagram shows a 45° prism made of glass and a ray of light entering the prism from air. Copy the diagram and show the path of the ray through the prism and back ‘i . ray of light into the air. e (ii) Explain why the ray follows the path that you have shown. 16. 45° prisu [2003 OL][2005 OL ]Give two uses of total internal reflection. 11 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. Optical fibres [2009 OL] Draw a diagram to show the path of a ray of light travelling through an optical fibre. [2009][2004][2004 OL] Explain how a signal is transmitted along an optical fibre. [2004 OL] How is the escape of light from the sides of an optical fibre prevented? [2004][2009] An optical fibre has an outer less dense layer of glass. What is the role of this layer of glass? [2004 OL] Give one use for optical fibres. [2004] Give two reasons why the telecommunications industry uses optical fibres instead of copper conductors to transmit signals. [2009] Impurities in glass reduce the power transmitted in an optic fibre by half every 2 km. The initial power being transmitted by the light is 10 W. What is the power being transmitted by the light after it has travelled 8 km through the fibre? [2009] Information is transmitted over long distances using optical fibres with a refractive index of 1.55. What is the speed of the light as it passes through the fibre? [2004] An optical fibre is manufactured using glass of refractive index of 1.5. Calculate the speed of light travelling through the optical fibre. [2004 OL] Name a material that is used in the manufacture of optical fibres. 12 20. Total internal reflection will only occur if the outer medium is of lesser density (strictly speaking it should read ‘lower refractive index’ rather than ‘less dense’, but there you go.) It also prevents damage to the surface of the core. 21. Endoscope, telecommunications, binoculars. 22. Less interference, boosted less often, cheaper raw material, occupy less space, more information carried in the same space, flexible for inaccessible places, do not corrode, etc. 23. After 2 km power has dropped to 5 W; after 4 km power has dropped to 2.5 W; after 6 km power has dropped to 1.25 W; after 8 km power has dropped to 0.625 W. 24, N= Can/Cglss Cys = Car/N Class = 3.0 * 10°/1.55 = Cylass = 1.94 x 10°m st 25. Ng= CalCg => 15=3x 107v,=> V, = 2.0 x 10°mst 26. Glass / plastic / sand / silicon Mandatory Experiments 27. (i) As in diagram, plus a ray box and protractor. (ii) The angle of incidence and the angle of refraction. (iii) By using the formula n = sini = sinr. (iv) To increase the accuracy of the results. 28. (i) See diagram. Also include a protractor and raybox. (ii) Draw the incident ray going in, the emergent ray coming out, then remove the block and join the two lines. This represents the refracted ray. (iii) By measuring the angle between the normal and the refracted ray using a protractor. (iv) angle of incidence i angle of refraction r sini sinr sin i/sinr | 30° 19° 0.500 0.326 1.53 45° 28° 0.707 0.469 1.51 65° 37° 0.906 0.602 1.50 (v) In each case sin i/sin r is (approximately) constant; therefore this verifies Snell’s Law. 29. (i) Diagram to show: A target medium e.g. glass block Incident ray (from ray box) Perpendicular / normal and refracted ray Label angles i andr (ii) Correct sin i and sin r values for six points Label axes correctly on graph paper sini | 0.500 | 0.643 | 0.76 | 0.819 | 0.866 | 0.906 | 0.939 Plot six points correctly 6 Straight line showing good distribution sinr | 0.325 | 0.438 | 0.50 | 0.544 | 0.588 | 0.615 | 0.643 (iii) A straight line through the origin shows that 0 sin i is proportional to sin r (iv) Correct slope method (n=) 1.41 [range: 1.38 — 1.52] (v) To reduce the (percentage) error Elaboration e.g. difficult to measure /read angles, r <i, etc. 30. (i) See diagram, plus ray-box. Mark the position of the incident and exit rays and also the outline of the block. Remove the block then measure the angle between the refracted ray and the normal using a protractor. a sini | 0.34 |050]064 10.7 | 087] 0.94 7 (iii) sin | 0.24 | 0.33 | 0.44 . 0.59 | 0.64 | Refractive index = slope = yo—yi/X.—X1 n = 1.49 15 (iv) There would be a greater percentage error associated with measuring smaller angles. Common Test errors State the laws of refraction of light Don’t use the word “reflection” in your answer What is meant by the refractive index of a material? Include the phrase “as light travels from a vacuum into that medium” Explain how a signal is transmitted along an optical fibre. Give three separate points Experiment to verify Snell’s law Using the recorded data, draw a suitable graph It must be a graph of sin i against sin r {not i andr} Using your graph, find the refractive index You must calculate the slope of the graph by using two points which are clearly marked on the graph {so you can’t take points from the table unless they also happen to be on the line of best fit which you have drawn} 16