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P3 Physics AQA 2012

P3 Physics AQA 2012. What is likely to come up?. Expect a question on Centre of Mass as AQA usually put one on. The Eye was new in 2012, and eye structure was on 2013 paper. This year long/short sightedness or near vision?

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P3 Physics AQA 2012

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  1. P3 Physics AQA 2012

  2. What is likely to come up? • Expect a question on Centre of Mass as AQA usually put one on. • The Eye was new in 2012, and eye structure was on 2013 paper. This year long/short sightedness or near vision? • Refractive index was new in 2012 but was not examined in 2013. • There has traditionally always been a ray drawing question. Last year it was a converging lens with object beyond 2f. In June 2012, it was a diverging lens. • A question on explaining transformers or another electromagnetic device – possible long answer. • Expect long/difficult calculations as there are few equations in P3. Remember workings and units.

  3. X-Rays • Absorbed strongly by bone or dense tissue. Can penetrate easily through low density tissue. • CCD’s can digitise image. • CT scans • Wavelength ~ size of atom (1x10-10m). • Ionising radiation – Risks to health • X-Ray therapy

  4. X-ray radiographs

  5. CCD’s

  6. CT Scanners

  7. CT Scans

  8. Ultrasound • Above 20KHz. For imaging, typically 1-5MHz. • Transducer sends out short pulses and detects echoes. • Reflected at a density boundary. • S=d/t applies to the echo. • A-scans and B-scans. • Ultrasound therapy/kidney stones/cleaners.

  9. Ultrasound Imaging • Ultrasound - Sound waves (longitudinal) with frequencies greater than 20KHz. • Typical frequencies used in imaging 1 – 5 MHz • The transducer produces short pulses and acts as a receiver in between pulses.

  10. Ultrasound Pulses

  11. Building the Image A-scan. The peaks correspond to reflection from surfaces inside the body. Taller peak represent stronger reflections. B-Scan. The transducer is moved across the body and the return signals are stored electronically. The signal strength controls the brightness/colour of each part of the image displayed on a screen. Blood vessels inside the eye

  12. How is light reflected from a surface? • Angles of incidence, reflection and refraction are always measured between the ray and the normal. • The normalat a point on a mirror is perpendicular to the mirror. • For a light ray reflected by a mirror: the angle of incidence = the angle of reflection. • Real images are formed where rays of light cross so they can be produced on a screen. • Virtual images cannot be produced on a screen – they are where rays of light appear to have come from.

  13. Image is virtual, same size as object, same distance behind mirror as object is in front, Upright and laterally inverted (back to front).

  14. SPL: Alfred Pasieka What is refraction? • Refraction of light is the change of direction of a light ray when it crosses a boundary between two transparent substances. • If the speed is reduced, refraction is towards the normal (e.g. air to glass). • If the speed is increased, reflection is away from the normal (e.g. glass to air).

  15. Refraction • n = Refractive index • Snells Law – n1sinθ1= n2sinθ2 • n for air/free space ~1 • If the ray is travelling from air into another medium, n1 = 1 and the expression becomes n2 = sin θ1 • sinθ2

  16. The angle of refraction depends of the angle of incidence and the refractive index of the two media. Air n ~ 1 • Glass n ~ 1.5 • Water n ~1.33

  17. Total Internal Reflection • Ifn1> n2(from a more to less dense material), then the critical angle sin c = n2/n1. • If the angle of incidence is greater that c, then total internal reflection will occur. • TIR used in endoscopes/reflectors/optical fibres.

  18. Endoscopes Uses Total Internal Reflection A bundle of fibres send the light into the dark cavity A bundle of fibres receive the reflected light into a camera Glass fibres are flexible and transmit/ receive images near the speed of light

  19. Lenses • Use curved surfaces to refract light. • Can be converging(convex) or diverging(concave). • Have a focal point. The focal length is measured from the lens to the focal point. • Power = 1/focal length (units D) +ve for converging lens, -ve for diverging lens. • Parallel rays imply the object is at optical infinity. • Image is produced where rays converge or appear to be diverging from.

  20. What are converging and diverging lenses? • A converginglens makes parallel rays of lightconverge to a focus. The point where they arefocused is the principal focus of the lens. • A diverging lens makesparallel rays of light diverge(spread out). The pointwhere the rays appear tocome from is the principalfocus of the lens. Corbis V257 (NT)

  21. Image formation • Principal rays • In parallel emerges through/from focal point. • Through the centre undeviated. • In through/towards focal point emerges parallel. • Image can be real or virtual. • Image can be upright or inverted • Image can be magnified or reduced. • Magnification = Image size/Object size.

  22. What type of images do the lenses form? • A real image is formed by a converging lens if the object is between its principal focus and infinity (magnified if between F and 2F, minified if beyond 2F). • A virtual image is always formed by a diverging lens, and by a converging lens if the object is between the principal focus and the lens.

  23. How is a converging lens used in a camera? • A camera contains a converging lens that is used to form a real image of an object.

  24. Principal rays for a converging lens

  25. How is a converging lens used in a magnifying glass? • A magnifying glass is a converging lens that is used to form a virtual image of an object. Photo: S. Meltzer/Photolink/Photodisc 24 (NT)

  26. Principal rays for a diverging lens

  27. Concave lenses are used to correct short sight. They always produce upright virtual images

  28. The eye • Retina, Lens, Cornea, Pupil/Iris, Ciliary Muscle, Suspensory Ligaments. • The cornea and lens cause convergence of rays on the retina. • Accomodation – Lens becomes more curved to focus on near object. • Far point is infinity and near point is ~25cm for ‘normal’ eye. • The eye is similar to a camera with the film/CCD representing the Retina.

  29. Gross structure Sclera – Tough inelastic outer coating. Choroid – Vascular layer – Retina needs a rich blood supply. Retina – Photosensitive layer. Cornea – Main light-focussing structure. contributes ~2/3 refractive power of the eye. Crystalline lens – Allows the eye to accommodate – adjust so that it is focussed on near objects. Pupil – Like the aperture of a camera. Allows light to enter the eye. The Iris can dilate/constrict the pupil to adjust the size of the pupil. Optic Nerve – The region where the optic nerve leaves the retina. Fovea – Very highly concentration of cone cells at the centre of our visual field

  30. Vision defects • Long sight – The eye is too short or not powerful enough. Rays converge behind the retina. A Converging lens is used to correct. • Short sight – The eye is too long or too powerful. Rays converge in front of retina. A Diverging lens is used to correct. • The focal length (and therefore power) of a lens is determined by the refractive index and the curvature of its surfaces.

  31. Refractive power • The normal eye has a power of around 60 Dioptres (D). • ~2/3 of this power is provided by the Cornea. • The rest is provided by the lens. Simple Refractive errors Myopia (short-sightedness) Hyperopia (long-sightedness)

  32. Moments • Moment = Force x Perpendicular distance between pivot and line of action of force. • For a system in equilibrium, the total clockwise moment must equal the total anticlockwise moment. • Levers. Small force x large distance to produce a large force at small distance.

  33. What is a moment? • A moment is the turning effect of a force. • Moment = force  perpendicular distance from the pivot to the line of action of the force = Fd • F is the force in newtons. • d is the perpendicular distance from thepivot in metres. • The unit of a moment is newton metres (Nm).

  34. What can you say about the moments of the forces acting on an object in equilibrium (it isn’t turning)? Higher The sum of the anticlockwise moments about any point = the sum of the clockwise moments about that same point.

  35. Using levers • Calculating moments is important when you use levers. • The weight is called the load. • The force you apply to the crowbar is the effort. • The point about which the crowbar turns is the pivot. • Levers enable you to lift heavy loads with little effort.

  36. What is the centre of mass of an object? • The centre of mass of an object is the point where its mass may be thought to be concentrated. • When a suspended object is in equilibrium, its centre of mass is directly beneath the point of suspension. • The centre of mass of a symmetrical object is along the axis of symmetry.

  37. Tilt or topple? Higher • An object will tend to topple over if the line of action of its weight is outside its base so … • … bodies with a low centre of mass and a broad base are more stable than bodies with a high centre of mass and a narrow base. • You can increase the stability of an object by making its base wider and its centre of mass as low as possible.

  38. THE PENDULUM The period of a pendulum is the time for one complete swing. The period of the pendulum T(sec) increases as the length increases. It is unaffected by the mass of the bob The best way to determine the period is to time 10 complete swings and divide by10. The frequency f of the swings = 1/ T and T = 1/ f

  39. Hydraulics • Pressure = Force/Area • Liquids transmit pressure in all directions as they are virtually incompressible. • Hydraulic systems use cylinders with different cross-sectional areas to turn a small force into a large one (same pressure in each cylinder).

  40. HYDRAULIC PRESSURE PRESSURE IN A FLUID IS THE SAME ALL THROUGH P = F1 = F2 A1 A2 SO A SMALL FORCE ON A SMALL AREA GIVES A BIG FORCE ON A BIG AREA

  41. The small piston of this hydraulic lift has a cross sectional area of 3 cm2 and the large piston has an area of 200 cm2. What force F must be applied to the small piston to lift a load of 15,000 N on the large piston? Show your working. Pressure at large piston due to load = 15000 200 = 75N/cm2 The pressure at the small piston must also = 75N/cm2 For small piston, P =F/A, then F = PxA = 75x3 = 225N

  42. Acceleration towards the centre of curvature as velocity is dependent upon direction. • The resultant force causing acceleration is called a centripetal force. • Centripetal force acts towards the centre of curvature. • Centripetal force required increases as mass and speed increase and as radius decreases.

  43. How can a body moving at a steady speed be accelerating? Circular Motion • When it’s moving in a circle at constant speed! • The object accelerates continuously towards the centre of the circle. • The centripetal force on it increases as: – the mass of the object increases, – the speed of the object increases, – the radius of the circle decreases.

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