phys1002_sln06_sm2_2007

# phys1002_sln06_sm2_2007 - PHYS1002 TUTORIAL 6 SOLUTIONS...

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PHYS1002 TUTORIAL 6 SOLUTIONS Semester 2, 2007 1. REFLECTION AND REFRACTION: DIAMONDS The written paragraph should address the following points: 1) Diamonds have a high index of refraction of 2.42 which means that at a diamond air interface the critical angle is only 24.4º. Because of this a substantial amount of light that enters the diamond can undergo total internal reflection, exiting at the face of the diamond to create the sparkling effect (as shown in the figure below). 2) The cut of the diamond is important because the angles of the diamond facets influence the effective angle of incidence of light beams when are incident on the back surface of the diamond. This then influences to which extent the incoming light is reflected back out of the diamond as shown below. [Image from Diamond Deals http://www.diamondideals.com ] 3) The index of refraction of diamond depends on the frequency of the incident light and so different colours of light diffract at different angles and so end up exiting the diamond at different positions along the diamond face. The wavelength dependence of diamond’s index of refraction is shown below. [Image from Gemological Institute of America http://www.gia.edu/]

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2. REFLECTION AND REFRACTION: PENTA PRISMS A penta prism [Description and image from Ross Optical Industries catalogue http://gateway.rossoptical.com/] a) Since the incident ray which first enters the prism is vertical we can easily find the angle of incidence with the normal to the upper prism surface. With a little bit of geometry we see that the angle of incidence ° = ° - ° = 5 . 22 5 . 67 90 i θ . The angle of reflection is thus 22.5°, giving a total deviation of 45 degrees from the vertical for this reflection. The normal to face D is 67.5° to the vertical. This gives an angle of incidence for the second reflection of ° = ° - ° = 5 . 22 45 5 . 67 i , so the total deviation of the ray at this surface is 45° away from the vertical, and the resulting ray is traveling horizontally, parallel to face A. We are told in the question that face C is perpendicular to face A, so the ray will travel through it undeviated, exiting the prism at 90° to its entry angle. b)
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phys1002_sln06_sm2_2007 - PHYS1002 TUTORIAL 6 SOLUTIONS...

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