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### 5.0 light

1. 1. Physics Module Form 4 Chapter 5 - Light GCKL 2011 5.1 UNDERSTANDING REFLECTION OF LIGHTWhat light is? Is a form energy. Light travel in a straight line and high speed about 300,000 km s-1.How the light 1. The light ray that strikes the surface ofray reflected the mirror is called incident ray.by the surface 2. The light ray that bounces off from theof mirror? surface of the mirror is called reflected ray. 3. The normal is a line perpendicular to the mirror surface where the reflection occurs. 4. The angle between the incident ray and the normal is called the angle of incidence AO = incident ray ,i. OB = reflected ray 5. The angle between the reflected ray and i = angle of incident the normal is called the angle of r = angle of reflected reflection, r.What is the The Laws Of ReflectionLaw of 1. The incident ray, the reflected ray and the normal all lie in the same plane The angle ofReflection ? incident, i, is ….equal….. to the angle of reflection, r.Draw the raydiagram of the 1. Consider an object O placed in front of aplane mirror plane mirror. 2. Measure the distance between the object o and the mirror. 3. Measure the same distance behind the mirror and mark the position as the image. 4. Draw the diverging ray from a point on the image to the corner of the eye. The rays from the image to the mirror must be dotted to show that are virtual. 5. Finally, draw two diverging rays from the object to the mirror to meet the diverging rays from the image. 5-1
2. 2. Physics Module Form 4 Chapter 5 - Light GCKL 2011State the 1 laterally invertedcharacteristics 2. same size as the objectof the image 3. virtualformed by 4. uprightplane mirror 5.distance between image and mirror same as distance between object and mirror.What is meantby virtualimage? Image that …cannot………. be seen on a screen.What is meantby real image? Image that …...can…be seen on a screen. CURVED MIRRORS: Concave mirror Convex mirror f f r rState the 1.Light (diverged, converged) 1.Light (diverged, converged)differences 2.(virtual,real) principal focus 2. (virtual,real) principal focusbetween 3.PF = Focal length 3. PF= ….Focal length…concave mirror = Distance between the virtual = Distance between the real principal focusand convex principal focus and the pole of the and the pole of the mirror.mirror mirror.Common Refer to the diagrams above and give the names for the following:terminology ofreflection of 1.Centre of curvature ,C = The geometric centre of a hollow sphere of which the concave orlight on a convex mirror is a part.curved mirror 2.Pole of mirror, P = The centre point on the curved mirror 3.Radius of curvature ,r = CP = radius of the curvature 4.Focal length, f = The distance between the principle focus, F and the pole of the mirror, P 5.Object distance, u = Distance of object from the pole of the mirror, P 6.Object distance , v = Distance of image from the pole of the mirror,PConstructionRules forConcaveMirror Ray 1 Ray 2 Ray 3 5-2
3. 3. Physics Module Form 4 Chapter 5 - Light GCKL 2011 A ray parallel to the A ray through F is reflected A ray through C is reflected principle axis is reflected to parallel to the principle back along its own path. pass through F. axis.Image formed Using the principles of construction of ray diagram, complete the ray diagrams for each of theby concave cases shown below:mirror: u = object distance; v = image distance ; f = focal length ; r = radius of curvature Note: Point of intersection in the position of the image A u < f ( Object between F and P ) Characteristics of Application: image: 1.magnifying mirror 1.virtual 2.sharing mirror 2.upright 3. make-up mirror 3.magnified B u = f ( Object, O is at F ) Characteristics of Application: image: A reflector to 1.Image at infinity produce parallel beam of light such as a reflector in 1. torchlight 2.spotlight C f < u < 2f or f < u < r ( Object O is Characteristics of between F and C image: 1.magnified 2.real 3.inverted I D u = 2f or u = r ( Object ,O is at C) Characteristics of Application: image: 1.reflector in a 1.same size projector 2.real 3.inverted 5-3
4. 4. Physics Module Form 4 Chapter 5 - Light GCKL 2011 Eu > 2f or u > r ( Object, O is beyond C ) Characteristics of image: 1.diminished 2.real 3.inverted I F u =  ( Object ,O very far from the lens) Characteristics of Application: image: Used to view distant 1.diminished objects as in a 2.real reflecting telescope 3.inverted IConstructionRules forConcaveMirror Ray 1 Ray 2 Ray 3 A ray parallel to the A ray towards F is reflected A ray towards C is reflected principal axis is reflected as parallel to the principal back along its own path. if it came from F. axis.Image formed Using the principles of construction of ray diagram, complete the ray diagrams for each of theby concave cases shown below:mirror: u = object distance; v = image distance ; f = focal length ; r = radius of curvature A u < f ( Object between F and P ) Characteristics of Application: image: 1. Blind Conner 1.diminished mirror 2.virtual 2.Wide side view 3.upright mirror 5-4
5. 5. Physics Module Form 4 Chapter 5 - Light GCKL 2011 Check Yourself: Objective Question:1. Which of the following is true of the laws of 4. A boy stands in front of a plane mirror a distance reflection f light? 5 m . When the boy moves toward the mirror by 2 A The angle of incident is equal to the m , what is the distance between the boy and his angle of refraction new image? B The incident ray and the reflected ray are always perpendicular to each A 2m B 4m other. C 6m D 8m C The incident ray , the reflected ray E 10 m and the normal line through the point of incidence, all lie on the same plane. 5. An object is placed in front of a plane mirror.2. The diagram shows a single ray of light being Compare to the object, the image formed in the directed at a plane mirror. mirror is always A virtual B smaller 40° C bigger D three times as far away What are the angles of incidence and reflection? Angle of incidence Angle of 6. A light ray incident onto a plane mirror at an reflection angle of 50o A 40o 40o The characteristics of an image , formed by a B 40o 50o convex mirror for all positions of the object are C 50o 40o A diminished, real and inverted D 50 o 50o B magnified , real, and upright C diminished ,virtual and upright3. The diagram shows a ray of light from a small bulb D magnified , virtual and inverted strikes a plane mirror. 7. A concave mirror has a focal length 20 cm. What happen to the size of image when an object is placed at a distance of 40 cm in front of the mirror? A diminished B magnified C same size of object Where is the image of the bulb formed and its characteristic? A At P and virtual B At Q and real C At R and virtual 5-5
6. 6. Physics Module Form 4 Chapter 5 - Light GCKL 20118. The figure shows a candle placed in front of a Section A (Paper 2) concave mirror of focal length, f. Structure Question: 1. Diagram 3.1 shows a mirror at the corner of a shop. The image formed is A real, upright and magnified B real, inverted and diminished DIAGRAM 3.1 / RAJAH 3.1 C virtual, inverted and magnified D virtual, upright and diminished (a) Name the type of mirror shown in Diagram 3.1 Convex mirror ……………………………………………………..9. When an object is placed at a point 20 cm in [1 mark] front of a concave mirror, a real image of the same as the object is formed on a screen placed (b) Name one characteristic of the image formed next to the object. What is the focal length of the by the mirror. mirror? Upright / diminished / smaller / virtual …………………………………………………….. A 5 cm [1 mark] B 10 cm (c) Sketch a ray diagram to show how the image C 15 cm is formed. D 20 cm 1. Draw a parallel ray from the object that is10. Which of the following states the right reason for incident along a path parallel to the principal axis replacing a plane mirror are used as rear- view appears to go through the focal point mirrors in motor vehicles with a convex mirror ? 2. A radial ray that is incident through the centre A To shine the object of curvature, C of the curved mirror is reflected B To widen the field of view back along the incident path through point C C To produce a brighter image D To produce a sharper image 3. Determine the correct position of the imageAnswer: 1 C 2 D 3 A 4 C [3 marks] 5 A (d) What is the advantage of using this type of 6 C mirror in the shop? 7 A To increase the field of vision …………………………………………………………… 8 A [1 mark] 9 B 10 C 5-6
7. 7. Physics Module Form 4 Chapter 5 - Light GCKL 20112. Diagram 4.1 shows the image of a patient’s teeth Section B(Paper 2) seen in a mirror used by a dentist. Essay Question(20 marks) Diagram 7.1 shows two cars, P and Q , travelling in the opposite directions, passing through a sharp band. A mirror is placed at X . DIAGRAM 4.1 (a) Name the type of the mirror used by the dentist. Concave mirror ………………………………………… ……. [ 1 mark ] (b) State the light phenomenon that causes the DIAGAM 7.1 image of the teeth (a) Diagram 7.2 shows an incomplete ray Reflection of light diagram when a plane mirror is placed at X. ……………………………………………….......... [ 1 mark ] (c) State two characteristics of the image formed. Virtual, upright and magnified ……………………………………………………. [ 2 marks ] (d) In the diagram below, the arrow represents DIAGRAM 7.2 the teeth as the object of the mirror. Complete the ray diagram by drawing the (i) Complete the ray diagram in Diagram 7.2 required rays to locate the position of the image. [2 marks] 1. Two reflected rays are shown (diagram)[1 mark] 2. Angle of incidence = Angle of reflection (diagram) [ 3 marks] [1 mark] 5-7
8. 8. Physics Module Form 4 Chapter 5 - Light GCKL 2011(ii) State the light phenomenon involved in (a)(i). (ii) Complete the ray diagram in Diagram 7.3 Reflection ………………………………………………… [2 marks] [1 mark](iii) Based on your answer in (a)(i), state the problem experienced by the driver in car P. The driver in car P cannot see car Q // field of ……………………………………………………….. view very small ……………………………………………………….. [1 mark] (b) Diagram 7.3 shows an incomplete ray diagram 1. Two reflected rays are shown (diagram)[1 mark] when a curve mirror is placed at X to replace the 2. Angle of incidence = Angle of reflection (diagram) plane mirror in Diagram 7.2. The curve mirror is used [1 mark] to overcome the problem that occur in (a)(iii). (iii) Based on your answer in b(ii), how the curve mirror solved the problem in (a) (iii)? The convex mirror increase the field of view ……………………………………………………. [1 mark] ( C) The characteristics of the image formed by the curved mirror in Figure 7.3 is diminished, virtual and upright. (i) What happen to the characteristics of the image when the focal length of the curved mirror DIAGRAM 7.3 is increased? The driver in car P cannot see car Q // field of (i) Give the name of the curve mirror. ……………………………………………………….. Convex mirror view very small …………………………………[1 mark] ……………………………………………….. [1 mark] (ii) Give the reason for your answer in (c)(i). The characteristics of image of a convex …………………………………………………….. mirror not depends on the focal length ………………………………………………… [1 mark] 5-8
9. 9. Physics Module Form 4 Chapter 5 - Light GCKL 2010 5.2 UNDERSTANDING REFRACTION OF LIGHT The diagram shows the spoon bent when put inside the water.State the Refraction of lightphenomenonoccurs.How the Light travel from less dense medium which is air to denser medium (water), light will bephenomenon deviated near to the normal. Thus the spoon seems like bending after putting inside the water.occurs?Why light is It due to change in the velocity of light as it passes from one medium into another.refracted? Light travel more slowly in water (or glass) than in air. When a light beam passes from air into glass, one side of the beam is slowed before the other. This makes the beam ‘bend’.Three differentcases of refraction Case 1: Case 2: Case 3: i = 0 ,r = 0 i>r i<r 5-9
10. 10. Physics Module Form 4 Chapter 5 - Light GCKL 2010 When a ray of light crosses Ray is light passes from Ray of light passes from the boundry between two air(less dense) to glass(dense) to air(less different mediums at a right glass(denser). dense) angle or the incident ray parallel to normal,  ray is bent towards  ray bent away from  the ray is not bent the normal the normal  but the speed of  the speed of light  the speed of light light is different. decreases after increases after  The angles of entering the glass emerging from the incidence and glass. refraction are zero.State the Laws of The Laws Of RefractionRefraction When the light travel from one medium to another medium which has a different optical density: 1. Snell’s Law :The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant. 2. The incident ray, the normal and the refracted ray all lie in the same planeRefractive Index 1. When light travels between two mediums with different optical densities, it changes speed and bends. 2. The speed of light will decrease when it enters an optically denser medium and increases when it enters an optically less dense medium. 3. The angle of bending of light depends on the refractive index of the mediums and the angle of incidence ,i.How to define 1. Refractive index, n is defined as, Example:refractive index n = sin i The diagram shows a ray of light passing from sin r air to the block X. where n = Refractive index i = the angle in medium less dense r = the angle in denser medium  A material with a higher refractive index has a higher density.  The value of refractive index , n  1 Calculate the refractive index of the block X.  The refractive index has no units. Solution: n = sin 50° sin 40° = 1.2 5 - 10
11. 11. Physics Module Form 4 Chapter 5 - Light GCKL 2010 2. Example: n = speed of light in vacuum (air) The speed of light in vacuum is 3 x 108 ms-1 speed of light in medium and the speed of light in glass is 2 x 108 ms-1 . Determine the refractive index of glass. Solution: 0r n = va n = 3 x 108 ms-1 2 x 108 ms-1 vm = 1.5 3.Real Depth and Apparent Depth The refraction of light gives us a false impression of depth. Example: A) The fish in the pond appears to be closer to the surface than it actually is. n = Real depth , H Apparent depth, h The following terms are defined: Or (B) The apparent depth – a swimming pool Real depth,H = The distance of the real object, looks shallower than it really is. n=H O from the surface of the water. h Apparent depth, h= The distance of the image, I from the surface of the water. 5 - 11
12. 12. Physics Module Form 4 Chapter 5 - Light GCKL 2010 ( C) A straight object place in water looks bent at the surface. Explanation: 1.Rays of light from the object travel from water to air. 2.Water is a denser medium compared to air. 3. Therefore, rays of light refract away from the normal as they leave the water. The rays of light then enter the eyes of the observer. 4. So the object appears to be nearer the surface of the water.Experiment to investigate the relationship between the Experiment to investigate the relationship betweenangle of incidence and the angle of refraction. real depth and apparent depth.Hypothesis: Hypothesis:The angle of refraction increases as the angle of The apparent depth increases as the real depth increases.incidence increases. . Aim of the experiment :Aim of the experiment :To investigate the relationship between the angle of To investigate the relationship between apparent depthincidence and the angle of refraction. and the real depthVariables in the experiment: Variables in the experiment:Manipulated variable: Angle of incidence Manipulated variable: real depthResponding variable: Angle of refraction Responding variable: apparent depthFixed variable: Refractive index Fixed variable: Refractive indexList of apparatus and materials: List of apparatus and materials:Glass block, ray box, white paper protactor, power Pin, ruler, water, retort stand ,tall beakersupply . 5 - 12
13. 13. Physics Module Form 4 Chapter 5 - Light GCKL 2010Arrangement of the apparatus: Arrangement of the apparatus:The procedure of the experiment which include the The procedure of the experiment which include themethod of controlling the manipulated variable and method of controlling the manipulated variable andthe method of measuring the responding variable. the method of measuring the responding variable.The glass block is placed on a white paper. A pin is placed at the base of the beaker as object O.The outline of the sides of the glass block are traced on The another pin is clamped horizontally onto the retortthe white paper and labelled as ABCD. stand as image position indicator, IThe glass block is removed. The beaker is filled with water.The normal ON is drawn. By using a ruler ,the real depth of the pin is measured, H=By using a protractor , the angle of incidence, i , is 8.0 cmmeasured = 20°. The pin O is seen vertically above the surface of theThe glass block is replaced again on its outline on the water.paper. The position of pin I is adjusted until parallax errorA ray of light from the ray box is directed along between the pin O and the pin I is non- existent.incidence line. By using the ruler again ,the position of pin I is measuredThe ray emerging from the side CD is drawn as line PQ. as the apparent depth = hThe glass block is removed again. The experiment is repeated 5 times for the other value ofThe point O and P is joined and is drawn as line OP. the real depth of water, ,i.e. D=10 cm,12 cm,14 cm andThe angle of refraction, r is measured. 16 cm.The experiment is repeated 5 times for the other angles of Tabulate the data:incidence, i= 30° , 40°,50°, 60° and 70°. H/cmTabulate the data: h/cm Sin i Sin r Analysis the data:Analysis the data: Plot the graph h against HPlot the graph Sin r against Sin i 5 - 13
14. 14. Physics Module Form 4 Chapter 5 - Light GCKL 2010Check Yourself:Objective Question:1 When light travels from one medium to another, 4 Figure shows a light ray travelling from refraction take place. Refraction is caused by the medium R to medium S. change in the A amplitude of light rays B intensity of light rays C strength of light rays D velocity of light rays2 An observer cannot see the coin in an empty glass Which of the following is true? as shown in Figure(a). However , he can see the coin when the glass is filled with water as shown A The speed of light in medium R is larger than in Figure(b). the speed of light in medium S B The optical density of medium R is larger than the optical density of medium S C The refractive index of medium R is larger than the refractive index of medium S 5 The diagram shows a light ray directed into a glass block. Which is the angle of refraction? Figure (a) Figure (b) The observer can see the coin in Figure (b) due to A the total internal reflection of light B the refraction of llight C the reflection of light D the diffraction of light3 Which of the following is not caused by the refraction of light ? 6 A light ray travels from medium P to medium Q. A A fish in pond appears nearer to the surface of the water Which of the following diagrams correctly shows B The sunlight reaches to the earth in a curve the path of the light ? path [ Medium P denser medium and Medium Q less C A ruler appears to bend at the water surface. dense ] D The sea water appear in blue colour 5 - 14
15. 15. Physics Module Form 4 Chapter 5 - Light GCKL 2010 9 Which of the following formulae can be used to determine the refractive index of a medium? A Angle of incidence Angle of refraction B Apparent depth Real depth C Speed of light in vacuum Speed of light in the medium7 The diagram shows a light ray travels from liquid 10 The diagram shows a light ray travels from the air L to liquid M. into medium X. Which of the following diagrams correctly shows the path of the light ? What is the refractive index of medium X? [ Refractive index of liquid M > Refractive index A 0.85 of liquid L ] B 1.24 C 1.31 D 1.41 E 1.58 11 The diagram shows a light ray travels from the oil into the air.8 The diagram shows a light ray which travels from the air to the glass. What is the value of k? [ Refractive index of oil = 1.4 ] A 44.4o B 45.6o C 54.5o What is the refrective index of the glass? D 55.4o E 58.9o A Sin S B Sin P Sin Q Sin R 12 The diagram shows a light of ray travels from the air into a glass block. C Sin Q D Sin R Sin R Sin S 5 - 15
16. 16. Physics Module Form 4 Chapter 5 - Light GCKL 2010 15 The diagram shows a coin is put at the base of the beaker. The image of the coin appears to be 5 cm from the base of the beaker. What is the refractive index of the glass block? A 1.38 B 1.45 C 1.51 D 1.62 What is the refractive index of the liquid? E 1.74 A 8 B 5 13 8 C 11 D 13 5 813 The speed of light in the air is 3 x 108 ms-1 . E 19 What is The speed of light in a plastic block? 14 [ Refractive index of plastic = 1.2 ] A 1.0 x 108 ms-1 Answer: B 1.5 x 108 ms-1 C 2.0 x 108 ms-1 1 D 11 A D 2.5 x 108 ms-1 2 B 12 C E 3.0 x 108 ms-1 3 D 13 D 4 A 14 C14 The diagram shows a boy appearing shorter when 5 D 15 D he is in a swimming pool. The depth of the water 6 C 16 in the pool is 1.2 m. 7 B 17 [ Refractive index of water = 1.33 ] 8 C 18 9 D 19 10 A 20 What is the apparent depth of the pool? A 0.1 m B 0.3 m C 0.9 m D 1.1 m E 1.6 m 5 - 16
17. 17. Physics Module Form 4 Chapter 5 - Light GCKL 2010Section A(Paper 2)Structure Question: (C ) (i) Draw a Diagram of the light ray shown on1. The Diagram shows a side view of a water-filled diagram 3.1, meeting the water surface RS, and showaquarium RSTU. An electric lamp, surrounded by a its path after meeting the surface. [1 mark]shield with a narrow transparent slit, is immersed inone corner of the aquarium at U. The light ray fromthe slit shines on the water surface RS at an angle of40o as shown in diagram below. R S Water 40o 40oR S o Water 40 Aquarium Light ray Light ray U T U T DIAGRAM 3.1 ii. Calculate the angle that this new path makes with RS and label the angle. ma [2 [1 mark] (a) What is meant by refractive index of a Angle = 40o substance? (d) The lamp is then placed outside underneath Refractive index is an indication of the light the aquarium with the light striking to the bottom of bending ability of the medium / the aquarium as shown in Diagram 3.2. Draw the light ray on Diagram 3.2, after striking the aquarium. n = sin i sin r [1 mark] [1 mark] (b) If the refractive index of water is 1.33, calculate the critical angle for a ray travelling from water to air. water n = 1 sin c sin c = 1 1.33 c = 48.80 [ 2 marks] Light ray Lamp 5 - 17
18. 18. Physics Module Form 4 Chapter 5 - Light GCKL 20102. An observer is looking at a piece of coin at the bottom of a glass filled with water as shown in Diagram 3. He found that the image of the coin is 3. Figure(a) shows an object in a small pond. The nearer to the surface of the water. depth of the water in the pond is H. The image of the objet appears to be h from water surface. [ 2 m a r k s Figure(a) ] (a) State the relationship between H and h (a)(i) State a characteristic of image in When H increases, h increases/ H is directly Diagram 3. proportional to h Virtual/magnified .................................................................... [1 mark] (b) When H = 4.5 ]m and the refractive index of (ii) Name the science phenomenon water is 1.33, determine the value of h . involve in the observation above. [1 mark] n = Real depth , H Refraction Apparent depth, h 1.33 = 4.5 m (b) Explain why the image of the coin H appears nearer to the surface of the H = 3.38 m water. - Light ray travels from density to less density medium (c) What happen to value of h when the pond is - Refracted ray away from normal poured with water of refractive index 1.40 ? [2 marks] H decreases (c) On Diagram 3, complete the ray …………………………………………… diagram from the coin to the observers eye. [2 marks] -Draw refracted ray correctly - Draw ray from image to the observer [ 5 - 18
19. 19. Physics Module Form 4 Chapter 5 - Light GCKL 2010Section B (Paper 2)Essay Question ii. Observe Figure 4(a) and Figure 4(b) carefully.1. Figure 4(a) shows a pencil placed in a glass of Compare the common characteristics of the pencilwater. Figure 4(b) shows the appearance of print and the print before and after they are removedviewed from the top of a thick block of glass placed from the water and the glass block respectively.over it. Use a physics concept to explain the appearance of the pencil and the print in water and under the pencil Glass block glass block respectively. [5 marks] Answer: 1. The pencil appears bent when placed in water and the print appears raised when a thick block of glass is placed over it. water 2. The rays of light from the pencil are refracted away Figure 4(a) Figure 4(b) from the normal as they leave the water and enter the eye of the observer. These rays appear to come from a virtual image above the actual point. The pencil (a) i. Why does the pencil appear bent to our eyes? ,therefore , appears bent in the water. Why does the print appear raised? [1 mark] 3. Rays of light from the print below the glass are refracted away from the normal as they leave the glass Answer: and enter the aye of the observer . The writing, therefore, appears to be slightly raised. We can see the pencil and the print because the rays of light from the two objects reach our eyes. 4. Refraction of light is the physics concept involved. 5. Refraction of light is a phenomenon in which rays of light change direction when they pass from one medium to another medium of a different density. 5 - 19
20. 20. Physics Module Form 4 Chapter 5 - Light GCKL 2011 5.3 UNDERSTANDING TOTAL INTERNAL REFLECTIONWhat is meant by Total internal reflection is the complete reflection of light ray travelling from a denser medium tototal internal a less dense medium.reflection? Total: because 100% of light is reflected Internal: because it happens inside the glass or denser medium.What is meant by The critical angle, c, is defined as the angle of incidence (in the denser medium) when the anglecritical angle ,c? of refraction (in the denser medium), r is 90°.What are therelationshipbetween thecritical angle andtotal internalreflection ? 5 - 20
21. 21. Physics Module Form 4 Chapter 5 - Light GCKL 2011  When the angle of  The refracted ray  If the angle of incidence, i keeps on travels along the incidence is increasing, r too glass-air boundary. increased is increases  Angle of refraction, r increased further so = 90°. that it is greater than  And the refracted ray moves further away  This is the limit of the critical angle, the light ray that can (i > c): from the normal be refracted in air as - no refraction  And thus approaches the refracted in air - all the light is the glass- air cannot be any larger totally in the boundary. than 90°. glass  The angle of  This phenomenon is incidence in the called total internal denser medium at the reflection. limit is called the critical angle, c.State the twoconditions for 1. light ray enters from a denser medium towards a less dense medium.total internalreflection to occur 2. the angle of incidence in the denser medium is greater then the critical angle of the medium ( i > c)What are therelationshipbetween therefractive index, nand critical angle,c?What are the 1. Mirage  In hot days, a person traveling in aphenomena car will see an imaginary pool ofinvolving total water appearing on the surface ofinternal reflection? the road.  The layes higher up are cooler and denser.  Light ray from the sky travels from denser to less dense medium and 5 - 21
22. 22. Physics Module Form 4 Chapter 5 - Light GCKL 2011 will refracted away from the normal.  The angle of incidence increases until it reach an angle greater than the critical angle.  Total internal reflection occurs and the light is reflected towards the aye of the observer.  If the observer’s eye is in the correct position, he will see a pool of water(image of the sky) appearing on the road surface.  This is known as a mirage. 2. Rainbow  When sunlight shines on millions of water droplets in the air after rain, a multi coloured arc can be seen.  When white light from the sun enters the raindrops, it is refracted and dispersed into its various colour components inside the raindrops.  When the dispersed light hit the back of the raindrop, it undergoes total internal reflection.  It is then refracted again as it leaves the drop.  The colours of a rainbow run from violet along the lower part of the spectrum to red along the upper part. 5 - 22
23. 23. Physics Module Form 4 Chapter 5 - Light GCKL 2011Give some 1. The sparkling of a diamond  A diamond has a high refractiveexamples of index.application of totalinternal reflection.  The higher the refractive index, the smaller the critical angle.  A small critical angle means total internal reflection readily occurs.  Light is easily reflected inside the diamond.  In this way, more light will be confined within the diamond before refracting out into the air. 2. Periscope  The periscope is built using two right angled 45° made of glass. The critical angle of the prism is 42°.  The angle of incidence is 45° which is greater than the critical angle.  Total internal reflection occurs.  The characteristics of the image are: Virtual, upright, same size. Give the advantages of the prism periscope compared to mirror periscope. Answer: (i) The image is brighter because all the light energy reflected. (ii) The image is clearer because there are no multiple images as formed in a mirror periscope. 5 - 23
24. 24. Physics Module Form 4 Chapter 5 - Light GCKL 2011 3. Prism Binocular  A light ray experiences two total internal reflections at each prism.  So the final image in binoculars is virtual, upright and same size. What are the benefits of using prism in binoculars? (a) an upright image is produced. (b) The distance between the objective lens and the eyepiece is reduced. This make the binoculars shorter as compared to a telescope which has the same magnifying power. 4. Optical Fibres  The external wall of a fibre optic is less dense than the internal wall.  When light rays travel from a denser internal wall to a less dense external walls at an angle greater than the critical angle, total internal reflection occurs. Give the advantage of using optical fibres cables over copper cables. (1) they are much thinner and lighter. (2) a large number of signals with very little loss over great distances. (3) The signals are safe and free of electrical interference 5 - 24
25. 25. Physics Module Form 4 Chapter 5 - Light GCKL 2011 (4) It can carry data for computer and TV programmes.Check Yourself:Objective Question:1 A ray of red light travelling in glass strikes the [ Refractive index of medium X = 1.3 glass-air boundary . Some light is reflected and Refractive index of medium Y = 1.5 ] some is refracted. Which diagram shows the paths of the rays? 4 Which of the following shows total internal reflection?2 One of the diagram below shows the path of abeam of light that is incident on a water-air surface with angle of incidence greater than the critical angle. Which one is it? 5 The diagram shows light ray XO experiencing total internal reflection when travelling from the glass to air.3 Which of the following diagram correctly shows the total internal reflection of ray of light? Which statements about total internal reflection are correct? 5 - 25
26. 26. Physics Module Form 4 Chapter 5 - Light GCKL 2011 P -  is more than the critical angle of glass Q - The speed of light in the glass is higher than in air R - The refractive index of glass is greater than air A P and Q B P and R In which direction does the light move from ? C Q and R D P,Q and R A OQ B OR6 The diagram shows a semi-circular plastic block C OS is placed in a liquid. D OT 9 A ray of light incident on one side of a rectangular glass block. If the angle of refraction in the glass block is 40o , which one of the following diagrams best represents this ray? [ The critical angle of glass is 42o ] Which of the following is correct? A Density of the plastic block is less than density of the liquid B Refractive index of the plastic block is less than refractive index of the liquid C Critical angle of the plastic block is less than critical of the liquid D Angle of incidence is less than critical angle of the liquid7 The diagram shows a ray of light passing through medium M to medium N. Which of the following is correct? 10 The diagram shows a light ray, P, directed into a glass block. The critical angle of the glass is 42o. A The angle of reflection is 55o In which direction does the light move from point B The critical angle of medium M less than 35o Q? C Density of medium M less than the density of medium N8 The figure shows a ray of light PO traveling in a liquid strikes the liquid-air boundary. [ The critical angle of the liquid = 45o ] 5 - 26
27. 27. Physics Module Form 4 Chapter 5 - Light GCKL 2011 C the greatest angle of incidence in optically more dense medium D the greatest angle of incidence in optically less dense medium 14 Which of the following shows the correct critical angle , c of the semi- circular glass block ?11 The diagram shows a light ray , M, directed into a glass block. The critical angle of the glass is 42o. In which diagram does the light move from point O? 15 The diagram shows a light ray travelling from air into a plastic block with an angle of incidence ,X. What is the critical angle of the plastic?12 The figure shows a ray of light is incident in air to the surface of Prism A and B. 16 The diagram shows a light ray travelling from air into a glass prism. Which comparison is correct ? A Density of prism A < density of prismB B Critical angle of prism A < critical angle of prism B C Refractive index of prism A < refractive index of prism B13 The critical angle is What is the critical angle of the glass? A the smallest angle of incidence in optically more dense medium B the smallest angle of incidence in optically less dense medium 5 - 27
28. 28. Physics Module Form 4 Chapter 5 - Light GCKL 2011 A 40o B 50o C 60o D 70o 21 The diagram shows a cross- section of a fibre E 80o optic cable.17 The refractive index of water is 1.33. What is the critical angle of the water. A 44.5o B 46.9o C 48.8 o D 49.2o Which comparison is correct ? E 54.3o18 The refractive index of plastic block is 13 . A Density of P < density of Q 5 B Density of P >density of Q What is the value of the cosine of the critical C Density of P = density of Q angle of the plastic? A 5 B 12 12 13 C 13 D 5 12 13 E 13 Answer: 5 1 A 11 C19 The figure shows a ray of light AO traveling in 2 D 12 B medium X strikes the medium X-air boundary. 3 D 13 C [ The refractive index of medium X = 1.12 ] 4 D 14 C 5 B 15 D 6 C 16 B 7 B 17 C 8 D 18 D 9 A 19 C 10 C 20 A In which direction does the light move from O ? A OE B OD C OC D OB20 Which of the following not applies the principle of total internal reflection? A Prism binocular B Mirror periscope C Optical fibre D Road mirage 5 - 28
29. 29. Physics Module Form 4 Chapter 5 - Light GCKL 2011Section A (c) Name other optical device that applies the(Paper 2) phenomenon in (a)(i).Structure Question: [ 1 mark ] Prism periscope // prism binoculars // camera//1. Diagram 1 shows a cross-sectional area of an endoscope and etc. optical fibre which consist of two layers of glass with different refractive index. The glass which 2. Figure 4 shows a traveller driving a car on a hot forms the inner core, Y is surrounded by another day. The traveller sees a type of glass which forms the outer layer, X. puddle of water on the road a short distance ahead of him. Puddle of water Figure 4 DIAGRAM 1 (a) (i) Name the light phenomenon observed in a) Which part of the air is denser? optical fibre? Close to the sky / cool air Total internal reflection ………………………………………………………… ( 1 mark ) [ 1 mark ] b) Name a phenomena of light that always depends (ii) Compare the refractive index of outer on the air density when light travels from the sky to layer X and inner core Y. the earth before it reaches point X. The refractive index of Y is higher than the refractive index of X// Vice versa Refraction ………………………………………………………… [ 1 mark ] ( 1 mark ) c) i) What is the phenomenon occurring at point X (b) The refractive index of inner core Y is 2.10. Calculate the critical angle of the inner core Total internal reflection Y. ……………………………………………………..… (1 mark ) 1 1 Sin c = = = 0.4762 n 2 .1 ii) What is the puddle of water actually? c = 28.44o // 280 26’ The image of sky [2 marks] ………………………………………………………… ( 1 mark ) 5 - 29
30. 30. Physics Module Form 4 Chapter 5 - Light GCKL 2011d) Using the diagram above, explain how the traveller can see the puddle of water on the road. 1.Light from sky to the earth refracted 2. The light reach at a point X, total internal reflection occurred………………………………………………………… ( 2 marks )e) Name one optical instrument that uses the phenomenon in (d)Optical fibre………………………………………………………… Answer: ( 1 mark ) Glass prism Object3. Completing the ray diagram below, to show how a 45operiscope works: (critical angle of glass = 42o) Total internal reflection takes place because angle of incident > critical angle Glass object prism tctct Eye Eye e 5 - 30
31. 31. Physics Module Form 4 Chapter 5 - Light GCKL 2011 5.4 U N D E R S T A N D I N G L E N S E SIntroduction Lenses are made of transparent material such as glass or clear plastics. They have two faces, of which at least one is curved.Types of lenses (a) Convex lens, also known as converging (b) Concave lens, also known as diverging lens. lens. It is thicker at the centre of the lens. It is thinner at the centre of the lens. Convex lens Concave lensState thedifferencesbetween convexlens and concavelens When light ray which are parallel and close to When light rays are parallel to the principle the principle axis strikes on a convex lens, they axis fall on a concave lens., they are refracted are refracted and converge to a point, F on the and appear to diverge from the focal point on principle axis. This point is a focal point of the the principle axis. convex lens.Common 1. The focal point, F is a point on the principle axis where all rays are close and parallel to theterminology of axis that converge to it after passing through a convex lens, or appear to diverge from it afterreflection of light passing through a concave lens.on a curved mirror 2. The focal length, f is the distance between the focal point and the optical centre. 3. The optical centre, C is the geometric centre of the lens. It is the point through which light rays pass through without deviation. 4. The principle axis is the line passing through the optical centre, C. 5 - 31
32. 32. Physics Module Form 4 Chapter 5 - Light GCKL 2011Construction rules Rule 1:of convex lens A ray parallel to the principle axis is refracted through the focal point, f. Rule 2: A ray passing through the focal point is refracted parallel to the principle axis. Rule 3: A ray passing through the optical centre, C travels straight without bending. The point of intersection is the position of the image. The images formed by a convex lens depend on the object distance, u.Images form by Using the principles of construction of ray diagram, complete the ray diagrams for each of theconvex lens cases shown below: u = object distance; v = image distance ; f = focal length Note: Point of intersection in the position of the image A u < f ( Object between F and P ) Characteristics of Application: image: 1.magnifying glass 1.virtual spectacle 2.upright 2.lens for long- 3.magnified sightedness. 4.Same side as the object B u = f ( Object, O is at F ) Characteristics of Application: image: 1. to produce a 1.virtual parallel a parallel 2.upright beam of light , as in a 3.magnified spotlight, 4. Same side as the astronomical object telescope 5 - 32
33. 33. Physics Module Form 4 Chapter 5 - Light GCKL 2011 C f < u < 2f or f < u < r ( Object O is Characteristics of Application: between F and C image: 1.projector lens 1.real 2.photograph 2.inverted 3. enlarger 3.magnified 4.objective lens of 4. On apposite side microscope of the object. D u = 2f or u = r ( Object ,O is at C) Characteristics of Application: image: 1.photocopying 1.real machine 2.inverted 3.same size as the object 4. On the opposite side of the object Eu > 2f or u > r ( Object, O is beyond C ) Characteristics of Application: image: 1.magnifying mirror 1.real 2.sharing mirror 2.inverted 3. make-up mirror 3.magnified F u =  ( Object ,O very far from the lens) Characteristics of Application: image: 1.magnifying mirror 1.virtual 2.sharing mirror 2.upright 3. make-up mirror 3.magnified 5 - 33
34. 34. Physics Module Form 4 Chapter 5 - Light GCKL 2011Construction rulesof concave lens Ray 1: A ray parallel to the principle axis is refracted 1 as if it appears coming from the focal point, F which is located at the same side of the 2 incident ray. 3 Ray 2: A ray passing through the focal point is refracted parallel to the principle axis. Ray 3: A ray passing through the optical centre, C travels straight without bending. The point of intersection is the position of the image . The image formed by a concave lens are always : Virtual, upright and diminished.Image formed by Using the principles of construction of ray diagram, complete the ray diagrams for each of theconvex mirror: cases shown below: u = object distance; v = image distance ; f = focal length A u < f ( Object between F and P ) Characteristics of Application: image: 1. Blind Conner 1.diminished mirror 2.virtual 2.Wide side view 3.upright mirror (B ) f<u <2f (object at between f and 2f) Characteristics of Application: image: 1. Blind Conner 1.diminished mirror 2.virtual 2.Wide side view 3.upright mirror 5 - 34
35. 35. Physics Module Form 4 Chapter 5 - Light GCKL 2011Power of lenses Power of a lens = 1 . Focal length The unit of power of a lens is Dioptre (D) or m-1 or P = 1 @ P = 100 Convex lens : the power is taken to be f f( cm) positive Concave lens : the power is taken to be negative f = focal length Linear Magnification (m) :The linearmagnification , mdefine as: Where f = focal length u = object distance v = image distance m = linear magnificationLens Formula ho = object height hi = image heightSign Conventions Type of lenses Convex lens Concave lens Object distance ,u Always + Always + Object is always placed to the Object is always placed to the left of the lens left of the lens Image distance, v + if the image is real ( image + if the image is real ( image is formed on the right side of is formed on the right side of the lens. the lens - if the image is virtual - if the image is virtual ( image is formed on the left ( image is formed on the left side of the lens). side of the lens). Focal length, f Always + Always - Power of length, P Always + Always - Linear magnification, m Size of image ImI =1 Image and object are the same size ImI >1 Enlarged image ImI <1 Diminished image 5 - 35