Presentation is loading. Please wait.

Presentation is loading. Please wait.

Physics 1230: Light and Color Ivan I. Smalyukh, Instructor Office: Gamow Tower, F-521 Phone: 303-492-7277 Lectures: Tuesdays.

Similar presentations


Presentation on theme: "Physics 1230: Light and Color Ivan I. Smalyukh, Instructor Office: Gamow Tower, F-521 Phone: 303-492-7277 Lectures: Tuesdays."— Presentation transcript:

1 Physics 1230: Light and Color Ivan I. Smalyukh, Instructor Office: Gamow Tower, F-521 Email: ivan.smalyukh@colorado.edu Phone: 303-492-7277 Lectures: Tuesdays & Thursdays, 3:30 PM - 4:45 PM Office hours: Mondays & Fridays, 3:30 PM – 4:30 PM TA: Jhih-An Yang jhihan.yang@colorado.edu Class # 6

2 2 Physics 1230: Light and Color Lecture 6: Reading: Finish Chapter 2 HW due today; New HW assigned (due in 2 weeks);

3 3 Chapter 2 – Geometrical Optics 1.Shadows 2.Reflection 3.Refraction 4.Dispersion We are here Geometrical optics is the theory of RAYS (straight lines) and how they reflect and refract (bend). Lots of similarity to GEOMETRY of lines and triangles. Main Topics

4 4 Chapter 2 – Geometrical Optics 1.Shadows 2.Reflection a)Specular or diffuse b)Equal angle rule c)Mirror images, ray tracing 3.Refraction 4.Dispersion a.Point source or diffuse source b.Umbra and penumbra c.How tall is my shadow? d.Pinhole camera We are here

5 What can happen to incoming light Transmitted Reflected (including scattering) absorbed Or any partial combination of these things Glass Silver, water Black pavement

6 Specular or diffuse? Diffuse reflection (paper) Specular reflection (mirror) Diffuse transmission (wax paper) 6

7 Equal angle rule (specular reflection) 7  r ii Normal Mirror  i = angle of incidence  r = angle of reflection  i =  r is specular reflection A normal is a line perpendicular to the surface. Incident Ray Reflected Ray

8 Plasma frequency of silver Metals with mobile electrons can cancel out the light field in forward direction - only reflection Metals reflect all waves below a certain frequency This plasma frequency varies from metal to metal Silver reflects light waves at all visible frequencies Gold and copper have a yellow-brownish color because they reflect greens, yellows and reds but not blues or violets Plasma frequency of gold Plasma frequency of copper

9 What is a mirror? Since silver is such a good reflector a coating of silver on glass - a good (common) mirror If the silver coating is thin enough the mirror can be made to transmit 50% of the light and to reflect the other 50% –This is called a half-silvered mirror –A half-silvered mirror used with proper lighting can show objects on one side or the other of the mirror

10 Law of specular reflection of a ray from a mirror Mirror This angle = this angle The normal to the mirror is an imaginary line drawn perpendicular to it from where the incident ray hits the mirror Normal The ray from the light bulb is diffusely reflected off chin. We show one of the many rays coming off his chin hitting a mirror. The incident ray undergoes specular reflection off the mirror –Note the reflected ray Draw the normal to the mirror –The angle of incidence = the angle of reflection

11 How is an image produced in a mirror? Part 1: Ray-tracing To find out how Bob "sees" Alex by looking in the mirror we trace rays which obey the law of reflection Mirror Alex Bob looks at Alex's image

12 We interpret all rays coming into our eye as traveling from a fictitious image in a straight line to our eye even if they are reflected rays! The psychology of image interpretation Mirror Alex Bob looks at Alex's image Example: To find the location of his hair in the virtual image we extend any reflected ray from hair backwards

13 The meaning of a virtual image If we trace rays for every ray from every part of Alex which reflects in the mirror –we get a virtual image of the real Alex behind the mirror. Virtual image of Alex is behind mirror Mirror Alex –Alex's virtual image is the same size as the real Alex Bob looks at Alex's image Bob sees Alex's image in the same place when he moves his head

14 Let’s look at specular reflection

15 Ray reflection practice retroreflector

16 Retroreflectors Bike reflectors Roadside reflectors Measuring distance to moon?

17 Half-silvered mirror If the silver coating is thin enough the mirror can be made to transmit 50% of the light and to reflect the other 50% This is called a half-silvered mirror A half-silvered mirror used with proper lighting can show objects on one side or the other of the mirror Glass is like a 4% silvered mirror

18 Image in a mirror 1.If a point on the object is distance X in front of the mirror, the same point in the image appears to be distance X in back of the mirror, or X object = X image. 2.The image point is on the normal (extended) from the object to the mirror. 18 X object X image normal extended Mirror

19 Ray tracing: Draw the image, then the rays 19 X object X image Mirror First: draw rays from image to eyes Viewed from the side.

20 Ray tracing: Draw the image, then the rays 20 X object X image Mirror First: draw rays from image to eyes Second: draw rays from mirror to object  i =  r happens automatically using this method. Demo on board

21 21 X object X image Mirror The top ray goes to the top of the bottle. It is right side up.  i =  r happens automatically using this method. Right side up image?

22 22 X object X image Mirror (to do this drawing, the mirror must be extended) The top ray goes to the bottom of the bottle. It is upside down.  i =  r happens automatically using this method. Right side up image? Extension

23 23 Mirror (to do this drawing, the mirror must be extended)  i =  r happens automatically using this method. Bottle on its side Viewed from the side.

24 For simple (flat) mirrors the image location is therefore predictable without knowing where the observer's eye is and without ray-tracing Mirror

25 Periscope 25 mirror The image of the bottle in the lower mirror is: A)Inverted B)Not inverted C)Something else Original OBJECT

26 Periscope? 26 extension Original OBJECT The first IMAGE

27 Periscope? 27 extension The second IMAGE

28 Periscope? 28 The image of the bottle in the lower mirror is: A)Inverted B)Not inverted C)Something else

29 Mirror Alex Question: Where are the images of Alex in the 2 mirrors? a)At A only b)At B only c)At A and B only d)At C only e)At A, B and C Multiple mirrors - a virtual image can act as a real object and have its own virtual image Mirror AC B The virtual image at A acts as an object to produce the virtual image of C. It acts as an intermediate image. More precisely it is the red rays which reflect as green rays.

30 30 Is the writing reversed? AR (Two mirrors, viewed from above) A)YES B)NO

31 31 Is the writing reversed? AR ЯAЯA

32 32 Is the writing reversed? AR ЯAЯA extension

33 33 Is the writing reversed? AR ЯAЯA A)YES B)NO

34 34 Lec. 6: Ch. 2 - Geometrical Optics 1.Shadows 2.Reflection 3.Refraction 4.Dispersion We are here

35 35 Refraction 1.Index of refraction: n = c / v 2.Ray in water is closer to the normal 3.Total internal reflection 4.Mirages

36 Reflection of waves occurs where the medium of propagation changes abruptly Part of the wave can be transmitted into the second medium while part is reflected back –You can hear someone from outside the pool when you are underwater because sound waves are transmitted from the air through the water (with different speed in each). When light waves are incident on a glass slab they are mostly transmitted but partly reflected (about 4%)! Glass slab Is the speed of light in the glass slab the same as in the free space??? No.

37 How can reflection require that the speed of the wave changes? We thought the speed of light was always c = 3 x 10 8 m/s! The speed of an electromagnetic (EM) wave is constant (for every wavelength) in empty space! The speed of light is slower than c in glass, water and other transparent media –(Einstein showed that light can never travel faster than c) The speed of light in a medium is v = c/n, where n is a number larger than one called the index of refraction n = 1.5 for glass n = 1.3 for water n = 1.5 for vegetable oil Light is reflected and transmitted at a boundary because –When a light wave travels in a medium the electric field of the light jiggles the electrons in the medium. –This produces new electric fields which can cancel or add to the original light wave both in the forward and backward directions These are the transmitted and reflected light waves

38 MaterialRefractive Index Air1.0008 Water1.330 Glass1.5 Diamond2.417 Ruby1.760 Oil1.5 Refractive indices of different materials Can we see a glass rod immersed into the oil with the same refractive index? A. Yes B. No Demo

39 Got to here on Thursday 6/10/2010

40 Refraction is the bending of a ray after it enters a medium where its speed is different A ray going from a fast medium to a slow medium bends towards the normal to the surface of the medium A ray going from a slow medium to a fast medium bends away from the normal to the surface of the medium The speed of light in a medium is v = c/n, where n is a number larger than one called the index of refraction and c = 3 x 10 8 m/s n = 1.3 for glass n = 1.5 for water Hence, a ray going into a medium with a higher index of refraction bends towards the normal and a ray going into a medium with a lower index of refraction bends away from the normal Air (fast medium) Glass or water (slow) Normal Glass or water (slow) Normal Air (fast medium) n air < n water 1.0008 < 1.33 How about light going into a medium with exactly the same index of refraction?

41 Ray-bending together with our psychological straight-ray interpretation determine the location of images underwater The precise amount of bending is determined by the law of refraction (sometimes called Snell's law): n i sin  i = n t sin  t Here,  i = angle between incident ray and normal, and  t = angle between transmitted ray and normal n i and n t are the indices of refraction in the medium containing the incident ray and in the medium containing the transmitted ray Fig 2.49 Fisherman and fish incident ray transmitted ray normal image of fish for someone out of water fish In order to observe the fish from outside the water a transmitted ray must enter your eye. You will think it comes from a point obtained by tracing it backwards, Extend any 2 of the many many transmitted rays from the fish backwards to find the image of the fish (where they intersect). The location of that image will be the same for any observer outside of the water.

42 What we see and how different it can be from what it seems to be The woman will see the underwater part of body being a)Smaller than it really is; b)Much larger than it really is; c)Of natural size;

43 The woman will see the underwater part of body being a)Smaller than it really is; b)Much larger than it really is; c)Of natural size; What we see and how different it can be from what it seems to be

44 The boy will see the underwater part of body being a)Smaller than it really is; b)Much larger than it really is; c)Of natural size;

45 Total internal reflection is an extreme case of a ray bending away from the normal as it goes from a higher to a lower index of refraction medium (from a slower to a faster medium) Glass or water (slow) Normal Air (fast medium) Just below the critical angle for total internal reflection there is a reflected and a transmitted (refracted) ray Glass or water (slow) Normal Just above the critical angle for total internal reflection there is a reflected ray but no transmitted (refracted) ray Critical angle For the glass-air interface

46 Total internal reflection Show that the internal reflection is a consequence of the Snell’s law The precise amount of bending is determined by the law of refraction (sometimes called Snell's law): n i sin  i = n t sin  t Here,  i = angle between incident ray and normal, and  t = angle between transmitted ray and normal n i and n t are the indices of refraction in the medium containing the incident ray and in the medium containing the transmitted ray

47 What we see and how different it can be from what it seems to be If the critical angle condition is satisfied, will the boy see the part of body above water: a)yes; b)No. Extra Credit: Refractive index of water is 1.33; What is the critical angle for the case of air-water interface?


Download ppt "Physics 1230: Light and Color Ivan I. Smalyukh, Instructor Office: Gamow Tower, F-521 Phone: 303-492-7277 Lectures: Tuesdays."

Similar presentations


Ads by Google