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Why does the spoon appear broken?. Refraction 23 September 2015 Objectives Be able to select and apply Snell’s Law. HSW: AF2 – Understanding the applications.

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Presentation on theme: "Why does the spoon appear broken?. Refraction 23 September 2015 Objectives Be able to select and apply Snell’s Law. HSW: AF2 – Understanding the applications."— Presentation transcript:

1 Why does the spoon appear broken?

2 Refraction 23 September 2015 Objectives Be able to select and apply Snell’s Law. HSW: AF2 – Understanding the applications and implications of science Used before in: lesson 3 Will use again in: lesson 5 PLTS: Team workers – Work collaboratively with others. Used before in: lesson 3Will use again in: lesson 5 Keywords frequency, speed, time, distance, Hertz, sonar, A-Scan, echo

3 Outcomes ALL MUST be able to carry out a practical safely MOST SHOULD calculate the refractive index of glass SOME COULD be able to apply the refraction equation to different situations

4 Refraction revision What can you remember? In groups make a mind map of everything you can remember about refraction. –When does it happen? –Can you draw diagrams to describe it? –Explain what apparent depth is.

5 Revision: Refraction occurs when a wave changes speed as it passes from one region to another. This speed change usually causes the wave to change direction. Water waves slow down as they pass over from a deeper to a shallower region. Light slows down as it passes from air into glass, perspex or water.

6 Refraction experiment Typical results: angle of incidence / ° angle of refraction / ° deviation / ° 000 15105 301911 452817 603525 754035 No deviation occurs when the angle of incidence is zero. Increasing the angle of incidence increases the deviation.

7 Refraction of light at a plane surface (a) Less to more optical dense transition (e.g. air to glass) angle of incidence normal AIR GLASS angle of refraction Light bends TOWARDS the normal. The angle of refraction is LESS than the angle of incidence.

8 (b) More to less optical dense transition (e.g. water to air) angle of refraction angle of incidence normal WATER AIR Light bends AWAY FROM the normal. The angle of refraction is GREATER than the angle of incidence.

9 object at the bottom of a pool AIR WATER Why a pool appears shallow image normals observer

10 Complete the paths of the RED light rays: A B C D E F

11 The refraction equation When a light ray passes from one medium to another: n = sin i sin r where: i is the angle of incidence in the first medium r is the angle of refraction in the second medium n is a constant number called the refractive index. r i

12 An experiment to find the refractive index (n) of glass 1.Set up the equipment as shown in the diagram opposite 2.For an initial angle of incidence, i of 30º trace the path of the light ray. 3.Measure the angle refraction, r. 4.Calculate the refractive index using the formula: n = sin (i) / sin (r). 5.Repeat for a range of angles between 10º and 80º. 6.Calculate the average value of n. i/ºsin(i/º)r/ºsin(r/º)n Average n

13 Question 1 Calculate the refractive index when light passes from air to glass if the angle of incidence is 30° and the angle of refraction 19º. n = sin i / sin r = sin (30º) / sin (19º) = 0.500 / 0.326 refractive index, n = 1.53

14 Question 2 Calculate the angle of refraction when light passes from air to perspex if the angle of incidence is 50° and the refractive index, n = 1.50. n = sin i / sin r 1.50 = sin (50º) / sin (r ) becomes: sin (r ) = sin (50º) / 1.50 = 0.766 / 1.50 sin (r ) = 0.511 angle of refraction = 30.7º

15 Question 3 Calculate the angle of incidence when light passes from air to water if the angle of refraction is 20° and the refractive index, n = 1.33. n = sin i / sin r 1.33 = sin (i) / sin 20º becomes: sin (i) = 1.33 x sin (20º) = 1.33 x 0.342 sin (i) = 0.455 angle of incidence = 27.1º

16 Complete: medium 1medium 2nir airwater1.3350 o 35.2 o glassair0.6730 o 48.6 o waterglass1.1359.8 o 50 o airdiamond2.4050 o 18.6 o airunknown1.5350 o 30 o Answers


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