Edexcel IGCSE Physics: Reflection and Refraction PDF

Head to www.savemyexams.com for more awesome resources Edexcel IGCSE Physics Your notes Reflection & Refraction Contents Light & Sound Waves Reflection & Refraction Core Practical: Investigating Refraction...

Head to www.savemyexams.com for more awesome resources Edexcel IGCSE Physics Your notes Reflection & Refraction Contents Light & Sound Waves Reflection & Refraction Core Practical: Investigating Refraction Snell's Law Core Practical: Investigating Snell's law Total Internal Reflection Page 1 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Light & Sound Waves Your notes Light Visible light is a part of the Electromagnetic spectrum which means it is a transverse wave This is explained in Transverse & longitudinal waves Representing a transverse wave Light waves are transverse: the particles vibrate in a perpendicular direction to the energy transfer Light can undergo: Reflection Refraction Sound Sound waves are longitudinal waves Page 2 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources This is explained in Transverse & longitudinal waves Longitudinal waves are usually drawn as several lines to show that the wave is moving parallel to the direction of energy transfer Your notes Drawing the lines closer together represents the compressions Drawing the lines further apart represents the rarefactions Representing a longitudinal wave Longitudinal waves are represented as sets of lines with rarefactions and compressions Sound can also undergo: Reflection Refraction The reflection of a sound wave is called an echo Page 3 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Reflection & Refraction Your notes Reflection & refraction All waves, whether transverse or longitudinal, can be reflected and refracted Reflection occurs when: A wave hits a boundary between two media and does not pass through, but instead stays in the original medium In optics the word medium is used to describe a material that transmits light Media means more than one medium An example of reflection An identical image of the tree is seen in the water due to reflection Refraction occurs when: Page 4 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources A wave passes a boundary between two different transparent media and undergoes a change in direction An example of refraction Your notes Waves can change direction when moving between materials with different densities The law of reflection The law of reflection states that: Angle of incidence (i) = Angle of reflection (r) Angles are measured between the wave direction (ray) and a line at 90 degrees to the boundary called the normal The angle of the wave approaching the boundary is called the angle of incidence (i) The angle of the wave leaving the boundary is called the angle of reflection (r) An example of reflection in a plane mirror Page 5 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Ray diagram of the reflection of a wave in a mirror Ray diagrams Reflection ray diagrams When drawing a ray diagram an arrow is used to show the direction the wave is travelling An incident ray has an arrow pointing towards the boundary A reflected ray has an arrow pointing away from the boundary A diagram showing the law of reflection Page 6 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes The angle of incidence and angle of reflection are equal in the law of reflection Refraction ray diagrams The direction of the incident and refracted rays are also taken from the normal line The change in direction of the refracted ray depends on the difference in density between the two media: From less dense to more dense (e.g air to glass), light bends towards the normal From more dense to less dense (e.g. glass to air), light bends away from the normal When passing along the normal (perpendicular) the light does not bend at all A diagram of a ray refracted into and out of a glass block Page 7 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes How to construct a ray diagram showing the refraction of light as it passes through a rectangular block The change in direction occurs due to the change in speed when travelling in different substances When light passes into a denser substance the rays will slow down, hence they bend towards the normal The only properties that change during refraction are speed and wavelength – the frequency of waves does not change Different frequencies account for different colours of light (red has a low frequency, whilst blue has a high frequency) When light refracts, it does not change colour (think of a pencil in a glass of water), therefore, the frequency does not change Examiner Tips and Tricks When drawing ray diagrams for reflection: 1. A simple straight line with an arrow is enough to represent the wave You do not need to draw the wavefronts unless asked to do so! 2. Take care to draw the angle correctly If it is slightly out it won’t be a problem, but if there is an obvious difference between the angle of incidence and the angle of reflection then you will probably lose a mark! Page 8 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Practice drawing refraction diagrams as much as you can! It's very important to remember which way the light bends when it crosses a boundary: Your notes As the light enters the block it bends towards the normal line Remember: Enters Towards When it leaves the block it bends away from the normal line Remember: Leaves Away Don't forget to draw the arrows for the direction of the light rays and make sure they are drawn with a ruler and a sharp pointed pencil Page 9 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Core Practical: Investigating Refraction Your notes Core practical 4: investigating refraction Aim of the experiment To investigate the refraction of light using transparent rectangular blocks, semi-circular blocks and triangular prisms To review your understanding of refraction use the revision note Reflection & refraction Variables Independent variable = shape of the block Dependent variable = direction of refraction Control variables: Width of the light beam Same frequency / wavelength of the light Equipment list Equipment Purpose Ray Box To provide a narrow beam of light that can be easily refracted Protractor To measure the angles of incidence and refraction Sheet of Paper To mark the lines indicating the incident and refracted rays Pencil To draw the incident and refracted ray lines onto the paper Ruler To draw the incident and refracted ray lines onto the paper Page 10 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Perspex blocks (rectangular block, semi-circular To refract the light beam block & prism) Your notes Resolution of measuring equipment: Protractor = 1° Ruler = 1 mm Method Refraction experiment set up Apparatus to investigate refraction 1. Place the glass block on a sheet of paper, and carefully draw around the rectangular perspex block using a pencil 2. Switch on the ray box and direct a beam of light at the side face of the block 3. Mark on the paper: A point on the ray close to the ray box The point where the ray enters the block Page 11 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources The point where the ray exits the block A point on the exit light ray which is a distance of about 5 cm away from the block Your notes 4. Draw a dashed line normal (at right angles) to the outline of the block where the points are 5. Remove the block and join the points marked with three straight lines 6. Replace the block within its outline and repeat the above process for a ray striking the block at a different angle 7. Repeat the procedure for each shape of perspex block (prism and semi-circular) Results Consider the light paths through the different-shaped blocks Refraction experiment results with different media Refraction of light through different shapes of perspex blocks The final diagram for each shape will include multiple light ray paths for the different angles of incidences (i) at which the light strikes the blocks This will help demonstrate how the angle of refraction (r) changes with the angle of incidence Page 12 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Label these paths clearly with (1) (2) (3) or A, B, C to make these clearer Use the laws of refraction to analyse these results Your notes You can use the revision note Reflection & refraction to do this Evaluating the experiment Systematic Errors: An error could occur if the 90° lines are drawn incorrectly Use a set square to draw perpendicular lines Random Errors: The points for the incoming and reflected beam may be inaccurately marked Use a sharpened pencil and mark in the middle of the beam The protractor resolution may make it difficult to read the angles accurately Use a protractor with a higher resolution Safety considerations The ray box light could cause burns if touched Run burns under cold running water for at least five minutes Looking directly into the light may damage the eyes Avoid looking directly at the light Stand behind the ray box during the experiment Keep all liquids away from the electrical equipment and paper Examiner Tips and Tricks You may be asked questions on how to perform this refraction experiment in your exam. You may also be required to complete a table of results or deduce the path of a refracted ray. Page 13 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Snell's Law Your notes Snell's law The angles of incidence and refraction are related to the refractive index of a medium by an equation known as snell's Law: sin i n= sin r Where: n = the refractive index of the material i = angle of incidence of the light (°) r = angle of refraction of the light (°) 'Sin' is the trigonometric function 'sine' which is on a scientific calculator You can revise the concept of refraction using the revision notes Reflection & refraction A formula triangle can help rearrange the snell's law equation Page 14 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Snell's law formula triangle For more information on how to use a formula triangle refer to the revision note on Speed Refractive index The refractive index is a number which is related to the speed of light in the material (which is always less than the speed of light in a vacuum): speed of light in a vacuum refractive index, n = speed of light in material The refractive index is a number that is always larger than 1 and is different for different materials Objects which are more optically dense have a higher refractive index, eg. n is about 2.4 for diamond Page 15 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Objects which are less optically dense have a lower refractive index, eg. n is about 1.5 for glass Since the refractive index is a ratio, it has no units Your notes Worked Example A ray of light enters a glass block of refractive index 1.53 making an angle of 15° with the normal before entering the block. Calculate the angle it makes with the normal after it enters the glass block. Answer: Step 1: List the known quantities Refractive index of glass, n = 1.53 Angle of incidence, i = 15° Step 2: Write the equation for snell's law sin i n= sin r Step 3: Rearrange the equation and calculate sin (r) sin i sin r = n sin( 15) sin r = 1. 53 sin r = 0. 1692 Step 4: Find the angle of refraction (r) by using the inverse sin function r = sin – 1 (0. 1692) = 9. 7 = 10° Examiner Tips and Tricks Page 16 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources sin i i Important: in snell's law is not the same as. Incorrectly cancelling the sin terms is a sin r r Your notes common mistake! When calculating the value of i or r start by calculating the value of sin i or sin r. You can then use the inverse sin function (sin–1 on most calculators by pressing 'shift' then 'sine') to find the angle. One way to remember which way around i and r are in the fraction is remembering that 'i' comes before 'r' in the alphabet, and therefore is on the top of the fraction (whilst r is on the bottom). Page 17 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Core Practical: Investigating Snell's law Your notes Core practical 5: investigating snell's law Aims of the experiment To investigate the refractive index of glass, using a glass block Variables Independent variable = angle of incidence, i Dependent variable = angle of refraction , r Control variables: Use of the same perspex block Width of the light beam Same frequency / wavelength of the light Equipment Equipment list Equipment Purpose Ray Box To provide a narrow beam of light that can be easily refracted Protractor To measure the angles of incidence and refraction Sheet of Paper To mark the lines indicating the incident and refracted rays Pencil To draw the incident and refracted ray lines onto the paper Ruler To draw the incident and refracted ray lines onto the paper Perspex rectangle To refract the light beam Resolution of measuring equipment: Page 18 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Protractor = 1° Ruler = 1 mm Your notes Method Diagram of equipment set up Apparatus set-up to investigate Snell's Law 1. Place the glass block on a sheet of paper, and carefully draw around the block using a pencil 2. Draw a dashed line normal (at right angles) to the outline of the block 3. Use a protractor to measure the angles of incidence to be studied and mark these lines on the paper 4. Switch on the ray box and direct a beam of light at the side face of the block at the first angle to be investigated 5. Mark on the paper: A point on the ray close to the ray box The point where the ray enters the block The point where the ray exits the block A point on the exit light ray which is a distance of about 5 cm away from the block Page 19 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources 6. Remove the block and join the points marked with three straight lines 7. Replace the block within its outline and repeat the above process for a rays striking the block at the Your notes next angle An example results table Angle of incidence, i / ° Angle of refraction, r / ° 0 10 20 30 40 50 60 70 80 Analysis of results If the angles have been measured correctly, the paper should end up looking like this: A diagram showing how to measure the angles of incidence and refraction Page 20 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Snell's Law relates the angles of incidence and refraction This is covered in the Snell's law revision note Plot a graph of sin i on the y-axis against sin r on the x-axis The refractive index is equal to the gradient of the graph A graph of the results of snell's law experiment Page 21 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Evaluating the experiment Systematic Errors: Your notes An error could occur if the 90° lines are drawn incorrectly Use a set square to draw perpendicular lines Random Errors: The points for the incoming and reflected beam may be inaccurately marked Use a sharpened pencil and mark in the middle of the beam The protractor resolution may make it difficult to read the angles accurately Use a protractor with a higher resolution Safety considerations The ray box light could cause burns if touched Run burns under cold running water for at least five minute Looking directly into the light may damage the eyes Avoid looking directly at the light Stand behind the ray box during the experiment Keep all liquids away from the electrical equipment and paper Page 22 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Total Internal Reflection Your notes Total internal reflection Sometimes, when light is moving from a denser medium towards a less dense one, instead of being refracted, all of the light is reflected This phenomenon is called total internal reflection Total internal reflection (TIR) occurs when: The angle of incidence is greater than the critical angle and the incident material is denser than the second material Therefore, the two conditions for total internal reflection are: The angle of incidence > the critical angle The incident material is denser than the second material The angles of refraction, critical and total internal reflection The critical angle is different for different materials. Refraction occurs when the angle of incidence is less than the critical angle, and total internal reflection occurs when it is greater Total internal reflection is utilised in optical fibres e.g. endoscopes Page 23 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources prisms e.g. periscopes Optical fibres Your notes Total internal reflection is used to reflect light along optical fibres, meaning they can be used for communications endoscopes decorative lamps Light travelling down an optical fibre is totally internally reflected each time it hits the edge of the fibre Light travelling in an optical fibre Optical fibres utilise total internal reflection for communications Structure of an endoscope Page 24 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Page 25 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Endoscopes utilise total internal reflection to see inside a patient's body Prisms Prisms are used in a variety of optical instruments, including periscopes binoculars telescopes cameras Prisms are also used in safety reflectors for bicycles and cars, as well as posts marking the edges of roads A periscope is a device consisting of two right-angled prisms that can be used to see over tall objects Reflection of light in a periscope When light travels through a periscope, it totally internally reflects through prisms causing the light to reflect at right angles Page 26 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources The light totally internally reflects in both prisms Reflection of light by right-angled prisms Your notes Single and double reflection through right-angled prisms Examiner Tips and Tricks If asked to name the phenomena make sure you give the whole name – total internal reflection. Remember: total internal reflection occurs when going from a denser material to a less dense material and ALL of the light is reflected. If asked to give an example of a use of total internal reflection, first state the name of the object that causes the reflection (e.g. a right-angled prism) and then name the device in which it is used (e.g. a periscope) Critical angle As the angle of incidence is increased, the angle of refraction also increases until it gets closer to 90° When the angle of refraction is exactly 90° the light is refracted along the boundary Page 27 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources At this point, the angle of incidence is known as the critical angle c Changing the angle of incidence to obtain the critical angle Your notes As the angle of incidence increases it will eventually surplus the critical angle and lead to total internal reflection of the light When the angle of incidence is larger than the critical angle, the refracted ray is now reflected This is total internal reflection Worked Example A glass cube is held in contact with a liquid and a light ray is directed at a vertical face of the cube. The angle of incidence at the vertical face is 39° and the angle of refraction is 25° as shown in the diagram. The light ray is totally internally reflected for the first time at X. Page 28 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Complete the diagram to show the path of the ray beyond X to the air and calculate the critical angle for the glass-liquid boundary. Answer: Step 1: Draw the reflected angle at the glass-liquid boundary When a light ray is reflected, the angle of incidence = angle of reflection Therefore, the angle of incidence (or reflection) is 90° – 25° = 65° Step 2: Draw the refracted angle at the glass-air boundary At the glass-air boundary, the light ray refracts away from the normal Due to the reflection, the light rays are symmetrical to the other side Step 3: Calculate the critical angle The question states the ray is “totally internally reflected for the first time” meaning that this is the lowest angle at which TIR occurs Therefore, 65° is the critical angle Page 29 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Your notes Examiner Tips and Tricks If you are asked to explain what is meant by the critical angle in an exam, you can be sure to gain full marks by drawing and labelling the same diagram above (showing the three semi-circular blocks). Calculating critical angle The critical angle, c, of a material is related to its refractive index, n The relationship between the two quantities is given by the equation: 1 sin c = n This can also be rearranged to calculate the refractive index, n: 1 n= sin c This equation shows that: The larger the refractive index of a material, the smaller the critical angle Light rays inside a material with a high refractive index are more likely to be totally internally reflected Page 30 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Worked Example Your notes Opals and diamonds are transparent stones used in jewellery. Jewellers shape the stones so that light is reflected inside. Compare the critical angles of opal and diamond and explain which stone would appear to sparkle more. The refractive index of opal is about 1.5 The refractive index of diamond is about 2.4 Answer: Step 1: List the known quantities Refractive index of opal, no = 1.5 Refractive index of diamond, nd = 2.4 Step 2: Write out the equation relating critical angle and refractive index 1 sin c = n Step 3: Calculate the critical angle of opal (co) 1 sin( c o ) = = 0. 6667 1.5 c o = sin – 1 (0. 6667) = 41. 8 = 42° Step 4: Calculate the critical angle of diamond (cd) 1 sin( c d ) = = 0. 4167 2.4 c d = sin – 1 (0. 4167) = 24. 6 = 25° Step 5: Compare the two values and write a conclusion Total internal reflection occurs when the angle of incidence of light is larger than the critical angle (i>c) In opal, total internal reflection will occur for angles of incidence between 42° and 90° The critical angle of diamond is lower than the critical angle of opal (co>cd) This means light rays will be totally internally reflected in diamond over a larger range of angles (25° to 90°) Page 31 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers  Head to www.savemyexams.com for more awesome resources Therefore, more total internal reflection will occur in diamond hence it will appear to sparkle more than the opal Your notes Examiner Tips and Tricks When calculating the value of the critical angle using the above equation: First use the refractive index, n, to find sin(c) Then use the inverse sin function (sin–1) to find the value of c Page 32 of 32 © 2015-2025 Save My Exams, Ltd. · Revision Notes, Topic Questions, Past Papers