{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lectures_Lect25

# Lectures_Lect25 - Physics 212 Lecture 25 Physics 212...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 212 Lecture 25 Physics 212 Lecture 25, Slide 1 The speed of light in a medium is slower than in empty space: since 0 < Remember: is dielectric constant (capacitors) Physics 212 Lecture 25, Slide 2 vmedium = c / nmedium A ray of light passes from air into water with an angle of incidence of 30 degrees. Checkpoint 1a Which of the following quantities does not change as the light enters the water? A. wavelength B. frequency C. speed of propagation 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 Physics 212 Lecture 25, Slide 3 Reflection Physics 212 Lecture 25, Slide 4 Refraction: Snell's Law D sin 2 D sin 1 = c / n2 c / n1 n2 sin 2 = n1 sin 1 Physics 212 Lecture 25, Slide 5 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 6 A Same Principle works for Light !! x1 l1 D y1 y2 x2 l2 B Time from A to B : To find minimum time, differentiate t wrt x1 and set = 0 How is x2 related to x1? Setting dt/dx1 = 0 l l t= 1+ 2 = v1 v2 2 2 2 2 x1 + y1 x2 + y2 + v1 v2 dt x1 x2 dx2 = + 2 2 2 2 dx1 v1 x1 + y1 v2 x2 + y2 dx1 x2 = D - x1 dx2 = -1 dx1 sin 1 sin 2 = v1 v2 v = c/n x1 x - 2 =0 v1l1 v2l2 n1 sin 1 = n2 sin 2 Physics 212 Lecture 25, Slide 7 Let's start with a summary: from n1 to n2 from n1 to n2 Physics 212 Lecture 25, Slide 8 The path of light is bent as it passes from medium 1 to medium 2. Checkpoint 2a Compare the indices of refraction in the two media. A. n1 > n2 B. n1 = n2 C. n1 < n2 Snell's Law: n1sin1 = n2sin2 n decreases increases Physics 212 Lecture 25, Slide 9 Total Internal Reflection NOTE: n1 > n2 implies 2 > 1 BUT: 2 has max value = 90o !! 1 > c Total Internal Reflection Physics 212 Lecture 25, Slide 10 A light ray travels in a medium with n1 and completely reflects from the surface of a medium n2. Checkpoint 2b The critical angle depends on: A. n1 only B. n2 only C. both n1 and n2 c clearly depends on both n2 and n1 Physics 212 Lecture 25, Slide 11 Intensity Anything looks like a mirror if light is just glancing off it. If two materials have the same n then its hard to tell them apart. Physics 212 Lecture 25, Slide 12 Polarization 56.3o 1 + 2 = 90 Snell's Law: sin 2 = sin( 90 - 1 ) = cos 1 n2 sin 2 = n2 cos1 = n1 sin 1 n2 tan1 = n1 Physics 212 Lecture 25, Slide 13 A ray of light passes from air into water with an angle of incidence of 30 degrees. Checkpoint 1b Some of the light also reflects off the surface of the water. If the incident light is initially unpolarized, the reflected light will be A. unpolarized B. somewhat horizontally polarized C. somewhat vertically polarized o = horizontal Physics 212 Lecture 25, Slide 14 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 15 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 no water ?? A w water Snell's Law: ray bent away from normal going from water to air Physics 212 Lecture 25, Slide 16 A light is shining at the bottom of a swimming pool (shown in yellow in the figure). A person is standing at the edge of the pool. Checkpoint 3 Can the person standing on the edge of the pool be prevented from seeing the light by total internal reflection at the water-air surface? A. yes B. no 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. DRAW SOME RAYS Physics 212 Lecture 25, Slide 17 Example: Refraction at water/air interface Diver's illusion 97 Diver sees all of horizon refracted into a 97cone air = 90 sinwater = na n 1 sin90 = a = nw nw 1.33 water = 48.5 Physics 212 Lecture 25, Slide 18 "Does gravity affect reflection/refraction? Would the reflection of the ray depend on say where you are in space? ex: standing on jupiter etc?" "Gravity affects the path taken by light, but small effect: Einstein ring" Physics 212 Lecture 25, Slide 19 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: Lig h t is re fra c te d a t th e s urfa c e o f th e wa te r Strategy: De te rm ine th e a ng le o f re fra c tio n in th e wa te r a nd e xtra p o la te th is to th e b o tto m o f th e ta nk. Physics 212 Lecture 25, Slide 20 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 R in the water ? A) R = 28.3o Snell's Law: B) R = 32.1o nairsin(45) = nwatersin(R) sin(R) = nairsin(45)/nwater = 0.532 R = sin1(0.532) = 32.1o Physics 212 Lecture 25, Slide 21 C) R = 38.7o 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? What number on the ruler does the laser beam hit ? A) 31.4 cm B) 37.6 cm tan(R) = d/50 45o 50 cm R nwater = 1.33 0 d 20 40 60 80 100 R = 32.1o C) 44.1 cm d = tan(32.1) x 50cm = 31.4cm Physics 212 Lecture 25, Slide 22 FollowUp 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 Physics 212 Lecture 25, Slide 23 45o 45o 50 cm R nwater = 1.33 50 cm d = 31.4 cm d = 50 cm 80 100 0 20 40 60 80 100 0 R = 32.1o 20 40 60 45o 50 cm R 0 20 40 nwater = 1.33 d = 40.7 cm 60 80 100 Physics 212 Lecture 25, Slide 24 25 cm + (31.4/2) cm A monochromatic ray enters a slab with n1 = 1.5 at an angle b as shown b TOP More Practice n = 1 n1 = 1.5 BOTTOM b n = 1 (A) Total internal reflection at the top occurs for all angles b, such that sinb < 2/3 (B) Total internal reflection at the top occurs for all angles b, such that sinb > 2/3 (C) There is no angle b (0 < b < 90o) such that total internal reflection occurs at top. Snell's law: n1 sin 1 = n2 sin 2 nsin is "conserved" Ray exits to air with same angle as it entered !! Physics 212 Lecture 25, Slide 25 FollowUp A ray of light moves through a medium with index of refraction n1 and is incident upon a second material (n2) at angle 1 as shown. This ray is then totally reflected at the interface with a third material (n3). Which statement must be true? 1 n1 n2 n3 (A) (B) (C) n1 < n3 n3 < n1 n2 n3 n2 1 If n1 = n3 n1 n2 n3 1 Want larger angle of refraction in n3 n3 < n1 Physics 212 Lecture 25, Slide 26 ...
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern