Light Revision

LightBy Neil Bronks

Light is a form of energy

Crookes Radiometer proves light has energy

Turns in sunlight as the light heats the black sideLight travels in straight lines

Reflection-Light bouncing off objectIncident rayNormalReflected rayAngle of incidenceAngle of reflectionMirrorAngle of incidence = Angle of reflectionLaws of ReflectionThe 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 ImageAn image that is formed by the eyeCan not appear on a screen

dd

Real ImageRays really meetCan be formed on a screenF2FFPrincipal AxisPoleConcave MirrorObjectAll ray diagrams in curved mirrors and lens are drawn using the same set of rays.FFYou can draw any ray diagram by combining 2 of these raysThe only difference is where the object is based.Your turn Paper and pen (Ruler naturally)F2FRay Diagrams- Object outside 2F1/. Inverted2/. Smaller3/. RealFThe images can be formed on a screen so they are real.2F13Object at 2F1/. Inverted2/. Same Size3/. RealThe image is at 2FF2F14Object between 2F and F1/. Inverted2/. Magnified3/. RealThe image is outside 2FF2F15Object at FThe image is at infinityF2FObject inside F1/. Upright2/. Magnified3/. VirtualThe image is behind the mirrorFConvex Mirror1/. Upright2/. Smaller3/. VirtualThe image is behind the mirrorFConvex Mirror only one ray diagramThe image is behind the mirrorFUses of curved mirrorsConcave Mirrors Dentists MirrorsMake up mirrors

Convex MirrorSecurity MirrorsRear view mirrors

CalculationsUse the formulaF

uvf=focal lengthu=object distancev=image distanceExampleAn 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

1020V=20cm real

2020MagnificationWhat is the magnification in the last question?Well u=20 and v=20As

22m=1Image is same size

ExampleAn 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

ExampleAn 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

Using the formula

V=60/5cm =12cm VirtualThe minus is Because theMirror is convexQuestionsAn 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

uvLamp-box

CrosswireScreenConcave mirrorApproximate 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.Refraction

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

Refraction into glass or waterLight bends towards the normal due to entering a more dense mediumAIRWATERRefraction out of glass or waterLight bends away from the normal due to entering a less dense mediumRefraction through a glass blockLight bends towards the normal due to entering a more dense mediumLight bends away from the normal due to entering a less dense mediumLight slows down but is not bent, due to entering along the normal34Refraction through a glass blockAngle of Incidence=iAngle of Refraction=rLaws of REFRACTIONThe incident ray, refracted ray and normal all lie on the same planeSNELLS 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 rProving Snells LawirSin iSin rA straight line though the origin proves Snells law. The slope is the refractive index.Proving Snells LawirSin iSin rA straight line though the origin proves Snells law. The slope is the refractive index.LaserGlass BlockProtractorH/WLC Ord 2006 Q2Refractive IndexRatio of speeds

Real and Apparent DepthA pool appears shallower

CorkPinMirrorApparent depthPinImageWaterReal depthMEASUREMENT OF THE REFRACTIVE INDEX OF A LIQUID Finding No Parallax Looking DownPin atbottomPin reflectionin mirrorParallaxNo 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/ApparentRepeat 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 mediumReflected like a mirrorAngle i = angle rSnells Window

(From dense to less dense

Finding the Critical Angle(From dense to less dense)

1) Ray gets refracted4) Total Internal Reflection3) Ray still gets refracted (just!)2) Ray still gets refractedTHE CRITICAL ANGLESemi-Circular Block Expt and on the internet click here2012 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 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 . What would be the value of the angle 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.

Mirages

RainbowRain DropletsStrong sunlightObserver back to sun

Critical AngleVaries according to refractive index

Prism Question Class ChallengeDraw the path of the light and give its angles70on=1.5Uses 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 informationOptical 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 ReflectionThe 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 kit2004 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/WLC Ord 2003 Q7Focal PointFocal PointLenses

Two types of lenses

Converging LensDiverging Lens

Focal Length=fFocal Length=f2FFFOptical CentreRay Diagrams2FFF2FFFDrawing again!ChallengeDraw the 5 ray diagrams for the converging lens and the diagram for the diverging lens . Write 3 characteristics of each image.2FFF2FFF2FConverging Lens- Object outside 2FImage is1/. Real2/. Inverted3/. Smaller2FFF2FObject at 2FImage is1/. Real2/. Inverted3/. Same size2FFF2FObject between 2F and FImage is1/. Real2/. Inverted3/. MagnifiedFFObject at FImage is at infinityFFObject inside FImage is 1/. Virtual2/. Erect3/. Magnified

CalculationsUse the formula

uvf=focal lengthu=object distancev=image distance2FFF2F

=-120ExampleAn 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

4030

v3040-V=120cm virtual

12030MagnificationWhat is the magnification in the last question?Well u=30 and v=120As

41Image is largeruvLamp-box with crosswireLensScreenMEASUREMENT OF THE FOCAL LENGTH OF A CONVERGING LENS Show on OPTICAL BENCH1. 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/WLC Ord 2002 Q3

AccommodationThe 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 internetDiverging LensFFImage is 1/. Virtual2/. Upright3/. Smaller

ExampleAn 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

Using the formula

V=60/5cm =12cm VirtualThe minus is Because theDiverging lens

=-20ExampleAn 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

-6030

v30-60-V=20cm virtual

2030MagnificationWhat is the magnification in the last question?Well u=30 and v=20As

23Image is smallerSign Conventionf PositiveVeitherf PositiveVeitherf negativeVnegativef negativeVnegative

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 VisionSolution is a pair of goggles!

Water

Power of LensOpticians use power to describe lenses.P=

So a focal length of 10cm= 0.1m is written as P=10m-1A diverging lens with a negative focal length f=-40cm=-0.4mHas a power of P = -2.5m-1

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 Glasses Class ChallengeA 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=-4The focal length is = -0.25mThe 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 m1. A person with defective vision has a minimum power of 64 m1. Calculate the focal length of the lens required to correct this defect. What type of lens is used? Name the defect.H/WLC Higher 2002 Q12 (b)LC Higher 2003 Q3