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# Lect25 - Physics 212 30 20 10 0 Confused Avg = 3.2...

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Unformatted text preview: Physics 212 Lecture 25 40 30 20 10 0 Confused Avg = 3.2 Confident Physics 212 Lecture 25, Slide 1 Physics Exam 3 next Wednesday 4/28 at 7PM • Covers Lectures 18-25 (that is, through today’s lecture) • Note: Homework 13, which is due Thu 4/29, has 3 problems on reflection and refraction which will be covered on the exam. (The due date was moved to 4/29 since there are 2 problems on the homework that are on lenses). • Review for the exam on Tuesday 4/27 at 5PM here. Physics 212 Lecture 25, Slide 2 Physics Let’s start with a summary: from n2 to n1 Physics 212 Lecture 25, Slide 3 Physics The speed of light in a medium is slower than in empty space: since ε > ε0 vmedium = c / nmedium Physics 212 Lecture 25, Slide 4 Physics Preflight 2 BB A B C What about the wave must be the same on either side ??? Observers in both media must agree on the frequency of vibration of the molecules 60 50 40 30 20 10 0 Physics 212 Lecture 25, Slide 5 Physics Reflection Physics 212 Lecture 25, Slide 6 Physics Refraction: Snell’s Law Physics 212 Lecture 25, Slide 7 Physics Marching Band analogy for refraction of light (bending of light) Paved parking lot, v=2 mph Let’s take photographs every 5 seconds from above Conclude: Direction that marching band is moving has “bent” after entering muddy field Muddy field, v=1 mph Physics 212 Lecture 25, Slide 8 Physics Think of a day at the beach… What's the fastest path to the ball knowing you can run faster than you can swim? This one is better Not the quickest route… Physics 212 Lecture 25, Slide 9 Physics A x1 l1 y1 y2 x2 l2 B 1 Time from A to B = t = l1/v1 + l2/v2 = v1 dt dx2 x1 x2 = + 2 2 2 2 dx1 v1 x1 + y1 v2 x2 + y2 dx1 -1 l1 l2 1 x +y + v2 2 1 2 1 2 2 x2 + y2 x1 x2 dt =0= − dx1 v1l1 v2l2 x1 x =2 v1l1 v2l2 sin θ1 sin θ 2 = v1 v2 Recall that v=c/n n1 sin θ1 = n2 sin θ 2 Physics 212 Lecture 25, Slide 10 Physics Preflight 6 70 Snell’s Law: n1sinθ1 = n2sinθ2 n decreases ﬂ θ increases 60 50 40 30 20 10 0 Physics 212 Lecture 25, Slide 11 Physics Total Internal Reflection θ1 > θc Simulation Total Internal Reflection Physics 212 Lecture 25, Slide 12 Physics Preflight 8 70 60 50 40 θc clearly depends on both n2 and n1 30 20 10 0 Physics 212 Lecture 25, Slide 13 Physics Intensity Anything looks like a mirror if light is just glancing off it. If two materials have the same n If then its hard to tell them apart. Physics 212 Lecture 25, Slide 14 Physics Preflight 4 70 60 50 40 30 20 10 0 Physics 212 Lecture 25, Slide 15 Physics Polarization sin θ 2 = sin(90o − θ1 ) = cosθ1 n2 tan θ1 = Snell’s Law: n2 sin θ 2 = n2 cosθ1 = n1 sin θ1 n1 θ1 + θ 2 = 90o Physics 212 Lecture 25, Slide 16 Physics A ball sits in the bottom of an otherwise empty tub at the front of the room. Suppose N people sit high enough to see the ball (N = ). Physics 212 Lecture 25, Slide 17 Physics A ball sits in the bottom of an otherwise empty tub at the front of the room. Suppose N people sit high enough to see the ball (N = ). Suppose I fill the tub with water but the ball doesn’t move. Will more or less people see the ball? A) More people will see the ball B) Same # will see the ball C) Less people will see the ball BB ?? θA θw n o wa t er water Snell’s Law: ray bent away from normal going from water to air Physics 212 Lecture 25, Slide 18 Physics Preflight 10 BB The light would go out in all directions, so only some of it would be internally reflected. The person would see the light that escaped after being refracted. 70 60 50 40 30 DRAW SOME RAYS 20 10 0 Physics 212 Lecture 25, Slide 19 Physics Example: Refraction at water/air interface • Diver’s illusion 97º Diver sees all of horizon refracted into a 97°cone θ1 = 90 o 1 n1 n1 o sin θ 2 = sin 90 = = n2 n2 1.33 θ 2 = 48.5o Physics 212 Lecture 25, Slide 20 Physics Exercise A meter stick lies at the bottom of a rectangular water tank of height 50cm. You look into the tank at an angle of 45o relative to vertical along a line that skims the top edge of the tank. What is the smallest number on the ruler that you can see? 45o nwater = 1.33 50 cm 0 20 40 60 80 100 Conceptual Analysis: - Light is refracted at the surface of the water Strategy: - Determine the angle of refraction in the water and extrapolate this to the bottom of the tank. Physics 212 Lecture 25, Slide 21 Physics Exercise A meter stick lies at the bottom of a rectangular water tank of height 50cm. You look into the tank at an angle of 45o relative to vertical along a line that skims the top edge of the tank. What is the smallest number on the ruler that you can see? 45o nwater = 1.33 50 cm 0 20 40 60 80 100 If you shine a laser into the tank at an angle of 45o, what is the refracted angle θR in the water ? A) θR = 28.3o B) θR = 32.1o C) θR = 38.7o BB Snell’s Law: nairsin(45) = nwatersin(θR) sin(θR) = nairsin(45)/nwater = 0.532 θR = sin-1(0.532) = 32.1o Physics 212 Lecture 25, Slide 22 Physics Exercise A meter stick lies at the bottom of a rectangular water tank of height 50cm. You look into the tank at an angle of 45o relative to vertical along a line that skims the top edge of the tank. What is the smallest number on the ruler that you can see? 45o 50 cm θR nwater = 1.33 0 d 20 40 60 80 100 θR = 32.1o What number on the ruler does the laser beam hit ? A) 31.4 cm B) 37.6 cm tan(θR) = d/50 d = tan(32.1) x 50cm = 31.4cm C) 44.1 cm BB Physics 212 Lecture 25, Slide 23 Physics Follow-Up Follow A meter stick lies at the bottom of a rectangular water tank of height 50cm. You look into the tank at an angle of 45o relative to vertical along a line that skims the top edge of the tank. 45o 50 cm nwater = 1.33 0 20 40 60 80 100 If the tank were half full of water, what number would the laser hit? (When full, it hit at 31.4 cm) A) 25 cm B) 31.4 cm C) 32.0 cm D) 40.7 cm E) 44.2 cm BB Physics 212 Lecture 25, Slide 24 Physics 45o 45o 50 cm θR nwater = 1.33 50 cm d = 31.4 cm d = 50 cm 80 100 0 20 40 60 80 100 θR = 32.1o 0 20 40 60 45o 50 cm θR 0 20 40 nwater = 1.33 d = 40.7 cm 60 80 100 Physics 212 Lecture 25, Slide 25 Physics 25 cm + (31.4/2) cm More Practice A monochromatic ray enters a slab with n1 = 1.5 at an angle θb as shown θb TOP n=1 n1 = 1.5 BOTTOM θb n=1 BB (A) Total internal reflection at the top occurs for all angles θb, such that sinθb < 2/3 (B) Total internal reflection at the top occurs for all angles θb, such that sinθb > 2/3 (C) There is no angle θb (0 < θb < 90o) such that total internal reflection occurs at top. Snell’s law: n1 sin θ1 = n2 sin θ 2 nsinθ is “conserved” Ray exits to air with same angle as it entered !! Physics 212 Lecture 25, Slide 26 Physics Follow-Up A ray of light moves through a medium with index of refraction n1 and is incident upon a second material (n2) at angle θ1 as shown. This ray is then totally reflected at the interface with a third material (n3). Which statement must be true? θ1 n1 n2 n3 (A) n3 < n1 (B) n1 < n3 ≤ n2 (C) n3 ≥ n2 θ1 If n1 = n3 n1 n2 n3 BB θ1 Want larger angle of refraction in n3 n3 < n1 Physics 212 Lecture 25, Slide 27 Physics ...
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