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Self-Study Package for Total Internal Reflection Click here to continue!

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Instructions Download and print out the document “Supplementary WS TIR.doc” You should go through these slides and work on the worksheets concurrently. Hence it is suggested that you do your worksheet in front of your computer and run this presentation. You should complete each section in the worksheet, check your answer to each section (in this presentation) BEFORE moving on to the next section. E.g. do section 2.1 first on the worksheet, check the answer to section 2.1 in this presentation, then move on to do section 2.2 on the worksheet. Press “Esc” at any page to end presentation

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Answers Click on the buttons below for the answer to each section DO NOT see the answer before you attempt the question yourself first! Back to instructions

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Section 1.1 Angle of Incidence (i) = Angle of Reflection (ii)

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Section 2.1 Light bends away from the normal, when it is going from more dense to less dense.

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Section 2.2

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Section 3.1 (Fig 3.1) n = sin r / sin i (1.5) = sin r / sin (35.5) sin r = (1.5)[(sin (35.5)] = (note: 4 sig. fig.) r = sin -1 (0.8711) = 60.59° (note: 2 d.p.) = 60.6° (final answer: 1 d.p.) 60.6°

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Section 3.1 (Fig 3.2) n = sin r / sin i (1.5) = sin r / sin (38.8) sin r = (1.5)[(sin (38.8)] = r = sin -1 (0.9399) = 70.03° = 70.0° (1 d.p.) 70.0°

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Section 3.1 (Fig 3.3) n = sin r / sin i (1.5) = sin r / sin (41.0) sin r = (1.5)[(sin (41.0)] = (4 sig. fig) r = sin -1 (0.9841) = 79.77° = 79.8° (1 d.p.) 79.8°

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Section 3.2 The refracted ray is getting closer and closer to the horizontal edge of the block The angle of refraction r is approaching 90°

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Section 3.3 We want to find i, when r = 90° Recall, n = 1.5 n = sin r / sin i (1.5) = sin 90 / sin i sin i = (sin 90) / (1.5) = i = sin -1 (0.6667) = 41.81° = 41.8° (1 d.p.) 90° 48.1°

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Section 3.4 Critical Angle

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Section 3.5 refer to textbook page 231 (Pure Physics), or 189 (Science Physics) The angle of incidence in an optically denser medium for which the angle of refraction in the optically less dense medium is 90° [Please memorize]

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Section 3.6

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Section 3.7 For water (n = 1.33) sin c = 1/n = 1/(1.33) = c = sin -1 (0.7519) = 48.8° For diamond (n = 2.41) sin c = 1/n = 1/(2.41) = c = sin -1 (0.4149) = 24.5°

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Section 4.1 Total internal reflection occurs Demo:

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Section 4.2 [refer to pg 231 (Pure Physics), or pg 189 (Science Physics)] The complete reflection of a light ray inside an optically denser medium at its boundary with an optically less dense medium. [please memorize]

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Section 4.3 i = 45° c = 41.8° i > c, therefore total internal reflection occurs 45°

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Section 4.4 For both cases, reflection occurs, and i = r Note: n = sin i/sin r does NOT apply

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Section 4.5 1) Light ray going from more dense to less dense medium 2) Angle of incidence is greater than critical angle [please memorize this]

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Section 4.6 (Case A) sin c = 1/n sin c = 1/(1.34) = c = sin -1 (0.7463) = 48.3° i < c, hence total internal reflection does not occur. Light goes through and is refracted. n = sin r/sin i 1.34 = sin r/sin 45 sin r = (1.34)(sin 45) = r = sin -1 (0.9475) r = 71.4° 71.4°

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Section 4.6 (Case B) sin c = 1/n sin c = 1/(1.83) = c = sin -1 (0.5464) = 33.1° i > c, hence total internal reflection occurs i = r 35°

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Section 4.6 (Case C) from section 3.3, c = 41.8° i > c, but total internal reflection does NOT occur, as light is going from less dense to more dense medium. Light passes through and is refracted n = sin i/ sin r (1.5) = sin 45/sin r sin r = sin 45/(1.5) = r = sin -1 (0.4714) = 28.1° 28.1°

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The End Click on the home button below to go back to the question menu Click anywhere else to end presentation

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