Light By Neil Bronks

Light is a form of energy Crooke’s Radiometer proves light has energy Turns in sunlight as the light heats the black side

Primary and secondary colours A secondary colour is 2 primary colours mixed Complimentary are that combine to make white

Light travels in straight lines

Reflection-Light bouncing off object Angle of incidence = Angle of reflection Normal Incident ray Reflected ray Angle of incidence Angle of reflection Mirror

Laws of Reflection The angle of incidence ,i, is always equal to the angle of reflection, r. The incident ray, reflected ray and the normal all lie on the same plane.

Virtual Image An image that is formed by the eye Can not appear on a screen d d

Real Image Rays really meet Can be formed on a screen F 2F

Concave Mirror Object Pole F Principal Axis All ray diagrams in curved mirrors and lens are drawn using the same set of rays. Concave Mirror Object Pole F Principal Axis

F

F You can draw any ray diagram by combining 2 of these rays The only difference is where the object is based. F

Your turn Paper and pen (Ruler naturally) 2F F

Ray Diagrams- Object outside 2F 1/. Inverted 2/. Smaller 3/. Real 2F F The images can be formed on a screen so they are real.

Object at 2F 1/. Inverted 2/. Same Size 3/. Real F 2F The image is at 2F

Object between 2F and F 1/. Inverted 2/. Magnified 3/. Real 2F F The image is outside 2F

Object at F F 2F The image is at infinity

Object inside F 1/. Upright 2/. Magnified 3/. Virtual F The image is behind the mirror

Convex Mirror 1/. Upright 2/. Smaller 3/. Virtual The image is behind the mirror 1/. Upright 2/. Smaller 3/. Virtual F

Convex Mirror – only one ray diagram F The image is behind the mirror

Uses of curved mirrors Concave Mirrors Convex Mirror Security Mirrors Dentists Mirrors Make –up mirrors Convex Mirror Security Mirrors Rear view mirrors

u v Calculations Use the formula f=focal length u=object distance v=image distance Use the formula u F v

Example An object is placed 20cm from a concave mirror of focal length 10cm find the position of the image formed. What is the nature of the image? Collect info f=10 and u=20 Using the formula 10 20 V=20cm real

Magnification What is the magnification in the last question? Well u=20 and v=20 As 20 m=1 Image is same size 2

Example An object is placed 20cm from a concave mirror of focal length 30cm find the position of the image formed. What is the nature of the image? Collect info f=30 and u=20 Using the formula V=60cm Virtual

Example An object is placed 30cm from a convex mirror of focal length 20cm find the position of the image formed. What is the nature of the image? Collect info f=-20 and u=30 The minus is Because the Mirror is convex Using the formula V=60/5cm =12cm Virtual

Questions An object 2cm high is placed 40cm in front of a concave mirror of focal length 10cm find the image position and height. An image in a concave mirror focal length 25cm is 10cm high if the object is 2cm high find the distance the object is from the mirror.

MEASUREMENT OF THE FOCAL LENGTH OF A CONCAVE MIRROR Crosswire u Lamp-box Screen v

Approximate focal length by focusing image of window onto sheet of paper. Place the lamp-box well outside the approximate focal length Move the screen until a clear inverted image of the crosswire is obtained. Measure the distance u from the crosswire to the mirror, using the metre stick. Measure the distance v from the screen to the mirror. Repeat this procedure for different values of u. Calculate f each time and then find an average value.   Precautions The largest errors are in measuring with the meter rule and finding the exact position of the sharpest image.

H/W 2007 H q3

Refraction The fisherman sees the fish and tries to spear it Fisherman use a trident as light is bent at the surface

Refraction into glass or water Light bends towards the normal due to entering a more dense medium AIR WATER

Refraction out of glass or water Light bends away from the normal due to entering a less dense medium

Refraction through a glass block Light bends towards the normal due to entering a more dense medium Light slows down but is not bent, due to entering along the normal Light bends away from the normal due to entering a less dense medium

Refraction through a glass block Angle of Incidence=i Angle of Refraction=r

Laws of REFRACTION The incident ray, refracted ray and normal all lie on the same plane SNELLS LAW the ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant for 2 given media. sin i = n (Refractive Index) sin r

Proving Snell’s Law Sin i i r Sin r A straight line though the origin proves Snell’s law. The slope is the refractive index.

Proving Snell’s Law Sin i i r Sin r Laser Sin i i Protractor Glass Block r Sin r A straight line though the origin proves Snell’s law. The slope is the refractive index.

H/W LC H 2010 Q3

Refractive Index Ratio of speeds

Real and Apparent Depth A pool appears shallower

REFRACTIVE INDEX OF A LIQUID MEASUREMENT OF THE REFRACTIVE INDEX OF A LIQUID Cork Pin Apparent depth Mirror Real depth Water Image Pin

Finding No Parallax – Looking Down Pin at bottom Pin reflection in mirror No Parallax Parallax

     Set up the apparatus as shown.     Adjust the height of the pin in the cork above the mirror until there is no parallax between its image in the mirror and the image of the pin in the water.     Measure the distance from the pin in the cork to the back of the mirror – this is the apparent depth.     Measure the depth of the container – this is the real depth.     Calculate the refractive index n= Real/Apparent Repeat using different size containers and get an average value for n.

Refraction out of glass or water (From dense to less dense) Light stays in denser medium Reflected like a mirror Angle i = angle r

Snell’s Window (From dense to less dense

Finding the Critical Angle…(From dense to less dense) 2) Ray still gets refracted 1) Ray gets refracted 4) Total Internal Reflection 3) Ray still gets refracted (just!) THE CRITICAL ANGLE

Semi-Circular Block Expt and on the internet click here

Mirages

Rainbow Rain Droplets Strong sunlight Observer back to sun

Critical Angle Varies according to refractive index

2012 Question 12 (b) [Higher Level] The diagram shows a ray of light as it leaves a rectangular block of glass. As the ray of light leaves the block of glass, it makes an angle θ with the inside surface of the glass block and an angle of 30o when it is in the air, as shown.   If the refractive index of the glass is 1.5, calculate the value of θ. What would be the value of the angle θ so that the ray of light emerges parallel to the side of the glass block? Calculate the speed of light as it passes through the glass.

Prism Question – Class Challenge Draw the path of the light and give its angles 70o n=1.5

Uses of Total Internal Reflection Optical fibres: An optical fibre is a long, thin, transparent rod made of glass or plastic. Light is internally reflected from one end to the other, making it possible to send large chunks of information Optical fibres can be used for communications by sending e-m signals through the cable. The main advantage of this is a reduced signal loss. Also no magnetic interference.

Practical Fibre Optics It is important to coat the strand in a material of low n. This increases Total Internal Reflection The light can not leak into the next strand.

Endoscopes (a medical device used to see inside the body): 2) Binoculars and periscopes (using “reflecting prisms”)

Now is a good time to get out the light demo kit

2004 Question 12 (b) [Higher Level] Give two reasons why the telecommunications industry uses optical fibres instead of copper conductors to transmit signals. Explain how a signal is transmitted along an optical fibre. An optical fibre has an outer less dense layer of glass. What is the role of this layer of glass? An optical fibre is manufactured using glass of refractive index of 1.5. Calculate the speed of light travelling through the optical fibre.

H/W LC Ord 2003 Q7

Lenses Converging Lens Diverging Lens Two types of lenses Focal Point Focal Length=f Focal Length=f Converging Lens Diverging Lens

Two refractions

Ray Diagrams Optical Centre 2F F F

2F F

Drawing again! 2F F

Challenge Draw the 5 ray diagrams for the converging lens and the diagram for the diverging lens . Write 3 characteristics of each image. 2F F

Converging Lens- Object outside 2F Image is 1/. Real 2/. Inverted 3/. Smaller 2F F

Object at 2F Image is 1/. Real 2/. Inverted 3/. Same size 2F F

Object between 2F and F Image is 1/. Real 2/. Inverted 3/. Magnified

Object at F Image is at infinity F

Object inside F Image is 1/. Virtual 2/. Erect 3/. Magnified F

u v Calculations Use the formula f=focal length u=object distance v=image distance Use the formula u 2F F v

Example An object is placed 30cm from a converging lens of focal length 40cm find the position of the image formed. What is the nature of the image? Collect info f=40 and u=30 Using the formula v 30 40 - = -120 40 30 V=120cm virtual

Magnification What is the magnification in the last question? Well u=30 and v=120 As 120 30 4 1 Image is larger

MEASUREMENT OF THE FOCAL LENGTH OF A CONVERGING LENS Show on OPTICAL BENCH Lamp-box with crosswire Screen Lens v u

1.      Place the lamp-box well outside the approximate focal length 2.    Move the screen until a clear inverted image of the crosswire is obtained. 3.    Measure the distance u from the crosswire to the lens, using the metre stick. 4. Measure the distance v from the screen to the lens.   5. Calculate the focal length of the lens using 6. Repeat this procedure for different values of u. 7. Calculate f each time and then find the average value.

H/W LC H 2009 Q2

Accommodation The width of the lens is controlled by the ciliary muscles. For distant objects the lens is stretched. For close up objects the muscles relax.

Accommodation internet

Diverging Lens Image is 1/. Virtual 2/. Upright 3/. Smaller F F

Example An object is placed 30cm from a diverging lens of focal length 20cm find the position of the image formed. What is the nature of the image? Collect info f=-20 and u=30 The minus is Because the Diverging lens Using the formula V=60/5cm =12cm Virtual

Example An object is placed 30cm from a diverging lens of focal length 60cm find the position of the image formed. What is the nature of the image? (Remember f must be negative) Collect info f=-60 and u=30 Using the formula v 30 -60 - = -20 -60 30 V=20cm virtual

Magnification What is the magnification in the last question? Well u=30 and v=20 As 20 30 2 3 Image is smaller

Sign Convention f Positive V either f negative V f negative V f

Myopia (Short Sighted) Image is formed in front of the retina. Correct with diverging lens.

Hyperopia (Long-Sighted) Image is formed behind the retina. Correct with a converging lens

Underwater Vision Solution is a pair of goggles! Water

Power of Lens Opticians use power to describe lenses. P=   So a focal length of 10cm= 0.1m is written as P=10m-1 A diverging lens with a negative focal length f=-40cm=-0.4m Has a power of P = -2.5m-1

The power of the total lens is Lens in Contact Most camera lens are made up of two lens joined to prevent dispersion of the light. The power of the total lens is Ptotal=P1+ P2 P = 1/12 -1/10=-0.0166cm-1 F=1/p=1/-0.0166=-60cm F=12cm F= -10cm

Glasses – Class Challenge A certain teacher has eyes of power 64m-1 but standard eyes are 60m-1 What is the power of the lens he needs and the focal length and type of the lens? The power is the difference 60-64=-4 The focal length is ¼ = -0.25m The minus means a diverging lens.

2006 Question 7 [Higher Level] What is meant by the refraction of light? A converging lens is used as a magnifying glass. Draw a ray diagram to show how an erect image is formed by a magnifying glass. A diverging lens cannot be used as a magnifying glass. Explain why. The converging lens has a focal length of 8 cm. Determine the two positions that an object can be placed to produce an image that is four times the size of the object? The power of an eye when looking at a distant object should be 60 m–1. A person with defective vision has a minimum power of 64 m–1. Calculate the focal length of the lens required to correct this defect. What type of lens is used? Name the defect.

H/W LC Higher 2002 Q12 (b) LC Higher 2003 Q3