#$&* course Phy 122 018. `Query 16*********************************************
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Given Solution: `a**** The angle of incidence and the angle of refraction are both measured relative to the normal direction, i.e., to the direction which is perpendicular to the surface. For a horizontal water surface, these angles will therefore be measured relative to the vertical. 66 degrees is therefore the angle of refraction. Using 1.3 as the index of refraction of water and 1 as the index of refraction of air, we would get • sin(angle of incidence) / sin(angle of refraction) = 1 / 1.3 = .7, very approximately, so that • sin(angle of incidence) = 1.3 * sin(angle of refraction) = .7 * sin(66 deg) = .7 * .9 = .6, very approximately, so that • angle of incidence = arcSin(.6) = 37 degrees, again a very approximate result. You should of course use a more accurate value for the index of refraction and calculate your results to at least two significant figures. STUDENT QUESTION it should be 1/1.333 right? nb is where its going which is air sin(66)/sin (theta)=1/1.333=.75 INSTRUCTOR RESPONSE The incident beam is in the water (call this medium a, consistent with your usage), the refracted beam in the air (medium b). It's sin(theta_a) / sin(theta_b) = n_b / n_a, so sin(theta_a) = 1 / 1.333 * sin(theta_b) = .7 sin(66 dec) = .6 and theta_a = 37 degrees. Again all calculations are very approximate. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qPrinciples of Physics and General College Physics 23.46 What is the power of a 20.5 cm lens? What is the focal length of a -6.25 diopter lens? Are these lenses converging or diverging? Your solution 1/.205m=4.88 diopters 1/-6.25= -16cm confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe power of the 20.5 cm lens is 1 / (.205 meters) = 4.87 m^-1 = 4.87 diopters. A positive focal length implies a converging lens, so this lens is converging. A lens with power -6.25 diopters has focal length 1 / (-6.25 m^-1) = -.16 m = -16 cm. The negative focal length implies a diverging lens. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!