PHY 171. Homework 5 solutions. (Due by beginning of class on Wednesday, February 8, 2012)


 Kelly Farmer
 2 years ago
 Views:
Transcription
1 PHY 171 (Due by beginning of class on Wednesday, February 8, 2012) 1. Consider the figure below which shows four stacked transparent materials. In this figure, light is incident at an angle θ on a boundary between two transparent materials. Some of the light then travels down through the next three layers of transparent materials, while some of it reflects upwards and escapes into the air. Calculate the values of (a) θ 5 and (b) θ 4? (a) Consider the n 1 to n 2 interface in the figure on the right. Since the angle of incidence at this interface is θ 1, the angle of reflection at the n 1 to n 2 interface is also θ 1 (as marked in the figure). From geometry (see figure on the right) the angle of incidence at the n 1 to n 5 interface will then also be θ 1. Note n 5 1, since it is air. Use Snell s law at the n 1 to n 5 interface: n 1 sin θ 1 n 5 sin θ 5 Given that θ , n , and n 5 1 (for air), we get: 1.30 sin sin θ 5 Solving this equation, we find that θ (b) To find θ 4, let us first set up the angles we will be using. Consider again the figure above, where at the n 1 to n 2 interface, we have designated the angle of refraction to be θ 2. Clearly, from geometry (see figure above), the angle of incidence for the n 2 to n 3 interface will then be equal to θ 2. Next, designate the angle of refraction at the n 2 to n 3 interface to be θ 3. Again from geometry, the angle of incidence for the n 3 to n 4 interface will be θ 3. Now, use Snell s law at each interface: At the n 1 to n 2 interface, Snell s law gives n 1 sin θ 1 n 2 sin θ 2, At the n 2 to n 3 interface, Snell s law gives n 2 sin θ 2 n 3 sin θ 3, At the n 3 to n 4 interface, Snell s law gives n 3 sin θ 3 n 4 sin θ 4, We don t need to calculate the intermediate angles θ 2 and θ 3, because things equal to the same thing are equal to one another. That is, we have from the above that so we can write n 1 sin θ 1 n 4 sin θ 4, n 1 sin θ 1 n 2 sin θ 2 n 3 sin θ 3 n 4 sin θ 4 from which we get 1.30 sin sin θ 4, so that θ
2 2. A point source of light is 80.0 cm below the surface of a body of water. Find the diameter of the circle at the surface through which light emerges from the water. If the angle of incidence is equal to the critical angle for the two media (water to air), then the refracted beam will just graze the surface of the water, as shown in the figure on the right. Therefore, light originating from the point source 80.0 cm below the surface that is within the cone whose vertex angle is equal to the critical angle will emerge from the water. On the other hand, light emerging at an angle greater than this will be totally internally reflected back into the water. Using Snell s law, 1.33 sinc 1 sin 90 since the angle of refraction is 90 when the angle of incidence is C, the critical angle. So, ( ) 1 C sin Retain for now more digits than significant to avoid rounding off errors. From geometry, C is also the vertex angle of the cone of light from the point source. This is clear from the figure above, where the twodimensional equivalent of this cone is shown. So with the vertex angle being C, the radius of the circle at the surface from which light emerges is given by r, which can be determined from trigonometry by setting tan C r 80.0 cm from which we obtain r (80.0 cm) tan Therefore, the required diameter of the circle is given by d 2r 2 (80 cm) tan( ) 182 cm 1.82 m Page 2 of 8
3 3. A diverging lens with a focal length of 15.0 cm and a converging lens with a focal length of 12.0 cm have a common central axis. The two lenses are separated by 12.0 cm. An object of height 1.00 cm is 10.0 cm in front of the diverging lens, on the common central axis. (a) Calculate where the lens combination produces the final image of the object. The setup is shown below. Use the lens equations one by one for each lens, keeping the sign convention (given in the Lecture notes) in mind. We will use d id as the image distance for the diverging lens with f d 15 cm. 1 For the diverging lens: 1 1 d o f d d id f d d o f d d o Therefore, d id f d d o d o f d ( 15) ( 15) d id 6 cm Since d id is negative, this means that the image is on the same side of the diverging lens as the object. This image now becomes the object for the converging lens. Since it is on the same side of the diverging lens as the object, we will need to add the distance between the lenses (D 12 cm) to the magnitude of this value (i.e., without considering the sign, since the sign only tells us where the image of the diverging lens alone is located). In other words, the object distance for the converging lens is d oc cm So please be careful, it is obvious that this is one of those problems where you will go wrong if you proceed blindly, without thinking about the signs, etc. Page 3 of 8
4 Moving along, if we use the lens equation again for the converging lens with f c +12 cm (i.e., a positive focal length this time since we are dealing with a convex lens), we get 1 d i 1 f c 1 d oc d oc f c f c d oc Therefore, d i f c d oc d oc f c (12) d i +36 cm This is the position of the final image. Since d i is positive, the image must to the right of the converging lens (i.e., on the opposite side from the diverging lens). (b) What is the height of the image? Show your calculations clearly if you want full credit. Again, proceed one by one. Object height h o 1.0 cm for the diverging lens, so m d d id d o ( 6) ; But m d h id 1.0 cm h id 0.6(1.0) 0.6 cm h id is then also the height of the object h oc for the converging lens, so m c d i (+36) 2; But m c h i d oc cm h i 0.6( 2) 1.2 cm Therefore, the height of the final image is 1.2 cm. The minus sign in the answer above indicates that the image is inverted, as we will discuss in part (d). (c) Is the image real or virtual? Explain. Since the final image formed by the lens combination is to the right of the converging lens (on the opposite side to the object), it must be formed by the light rays actually passing through that location, so it is a real image. (d) Does the image have the same orientation as the object or is it inverted? Explain. In part (b) above, we calculated the magnification for the lens combination, and found it to be negative. This means that the image is inverted. This is as expected the concave lens should keep its image upright, so the object for the convex lens is upright, and the convex lens will produce an inverted image. Page 4 of 8
5 4. In a doubleslit arrangement, the slits are separated by a distance equal to 100 times the wavelength of the light passing through the slits. (a) What is the angular separation in radians between the central maximum and an adjacent maximum? (a) The condition for a maximum in the 2slit interference pattern is d sinθ mλ where m 0, 1, 2,... is called the order of the maxima, and d is the slit separation. For the central maximum, m 0 in the above equation. For the first offcenter maximum, m 1. The angular separation θ between the central maximum and the first offcenter maximum is then given by d sin θ λ. In this experiment, we are told that d 100λ. So, sin θ λ d which gives θ radians λ 100λ (b) What is the difference between these maxima on a screen 50.0 cm from the slits? The geometry of the problem is shown in the figure below. From the figure, we see that for small θ, we get sin θ tanθ y/l Equating to the value of sinθ in part (a), we get y L so that y L 50.0 cm 0.50 cm Page 5 of 8
6 5. A doubleslit arrangement produces interference fringes for sodium light (λ 589 nm) that are 0.20 apart. What is the angular fringe separation if the entire arrangement is immersed in water (n 1.33)? Once again, the condition for a maximum in the 2slit interference pattern is d sinθ mλ where m 0, 1, 2,... is called the order of the maxima, and d is the slit separation. For the central maximum, m 0 in the above equation. For the first offcenter maximum, m 1. The angular separation θ between the central maximum and the first offcenter maximum is then given by d sin θ λ, which can be rewritten as from which we get d λ/sin d λ sin θ If immersed in water, the frequency remains the same but the speed of light in water changes to a smaller value v water. To find v water, recall that the index of refraction (n) of a medium is related to v water via the relation n c/v water, where c is the speed of light in air. So, putting v water c/n, we get the new wavelength when the arrangement is immersed in water to be But c/f λ, the wavelength in air. λ water v water f Therefore, the wavelength in water is given by c/n f where n 1.33 for water. λ water λ n So, we can write for the new angular fringe separation (θ water ) between the central maximum and the first offcenter maximum that d sin θ water λ water from which we get sin θ water λ water d λ/n sin 0.20 λ/sin 0.20 n sin from which we get the angular fringe separation in water to be θ water 0.15 Page 6 of 8
7 6. The distance between the first and the fifth minima of a singleslit diffraction pattern is 0.35 mm with the screen 40.0 cm away from the slit, when light of wavelength nm is used. (a) Find the slit width. We are given that the distance between the first and fifth diffraction minima is 0.35 mm, the distance from slit to screen is L 40 cm 400 mm, and the wavelength is λ 550 nm mm. Diffraction minima occur at where a is the slit width, and m 1, 2, 3,.... a sin θ m λ (P6.1) As in the interference setup, we have for small θ that sin θ tanθ y/l, where y is the distance from the central point on the screen (i.e., the middle of the central bright fringe) to the position of the minima on the screen. Substituting this relation in equation (P6.1 ), we get ( y a m λ L) from which we get y mlλ/a. So, for the 1 st minimum (m 1), y 1 Lλ a, whereas for the 5th minimum (m 5), y 5 5Lλ a. We are given that y 5 y mm, so So, a 0.35 mm y 5 y 1 5Lλ a Lλ a 4Lλ a 4Lλ 0.35 mm 4 (400 mm) mm 0.35 mm 2.5 mm Therefore, the width of the slit is 2.5 mm (b) Calculate the angle θ of the first diffraction minimum. For the first minimum, a sin θ 1λ. So sin θ λ a mm 2.5 mm This gives θ radians Page 7 of 8
8 7. In a doubleslit experiment, the slit separation d is 2.00 times the slit width w. How many bright interference fringes are there in the central diffraction envelope? We are given that the slit separation, d 2w, where w is the slit width. Let L be the perpendicular distance from the slits to the screen. Since the condition for diffraction minima is w sin θ mλ, where m 1, 2,... (with w here being just a different symbol for the more commonly used a), the first diffraction minima will occur for m 1 at an angle θ D such that sin θ D λ w Therefore, if the first diffraction minimum is at a distance y D from the central spot on the screen, we will get for small θ (for which sin θ D tan θ D y D /L) that ( ) λ y D L w Meanwhile, the condition for interference maxima is d sin θ mλ, where m 0, 1, 2,...; if the m th interference maximum is at a distance y I from the central spot on the screen, we will get for small θ (for which sin θ I tan θ I y I /L) that ( ) ( ) mλ mλ y I L L d 2w To find how many bright interference fringes are in the central diffraction envelope, all we need to do then is set y I y D, and solve for m. This gives ( ) ( ) mλ λ L L 2w w from which we obtain m 2. This means that the 2 nd interference maximum will coincide with the first diffraction minimum. If you look at the figure of how the diffraction envelope sits on top of the interference pattern in the lecture notes, you should realize that this means that we will not see any intensity at this position. So, the only interference fringes we can see within the central diffraction envelope are the central interference maximum, and the first offcenter interference maxima, one to either side of the central maximum. In summary, this means that there are 3 interference maxima visible within the first diffraction envelope. Page 8 of 8
AP Physics B Ch. 23 and Ch. 24 Geometric Optics and Wave Nature of Light
AP Physics B Ch. 23 and Ch. 24 Geometric Optics and Wave Nature of Light Name: Period: Date: MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) Reflection,
More informationChapter 23. The Refraction of Light: Lenses and Optical Instruments
Chapter 23 The Refraction of Light: Lenses and Optical Instruments Lenses Converging and diverging lenses. Lenses refract light in such a way that an image of the light source is formed. With a converging
More informationConvex Mirrors. Ray Diagram for Convex Mirror
Convex Mirrors Center of curvature and focal point both located behind mirror The image for a convex mirror is always virtual and upright compared to the object A convex mirror will reflect a set of parallel
More informationRutgers Analytical Physics 750:228, Spring 2016 ( RUPHY228S16 )
1 of 13 2/17/2016 5:28 PM Signed in as Weida Wu, Instructor Help Sign Out Rutgers Analytical Physics 750:228, Spring 2016 ( RUPHY228S16 ) My Courses Course Settings University Physics with Modern Physics,
More information1 of 9 2/9/2010 3:38 PM
1 of 9 2/9/2010 3:38 PM Chapter 23 Homework Due: 8:00am on Monday, February 8, 2010 Note: To understand how points are awarded, read your instructor's Grading Policy. [Return to Standard Assignment View]
More informationSolution Derivations for Capa #14
Solution Derivations for Capa #4 ) An image of the moon is focused onto a screen using a converging lens of focal length (f = 34.8 cm). The diameter of the moon is 3.48 0 6 m, and its mean distance from
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) A single slit forms a diffraction pattern, with the first minimum at an angle of 40 from
More informationPhysics 111 Homework Solutions Week #9  Tuesday
Physics 111 Homework Solutions Week #9  Tuesday Friday, February 25, 2011 Chapter 22 Questions  None MultipleChoice 223 A 224 C 225 B 226 B 227 B 229 D Problems 227 In this double slit experiment we
More information6) How wide must a narrow slit be if the first diffraction minimum occurs at ±12 with laser light of 633 nm?
Test IV Name 1) In a single slit diffraction experiment, the width of the slit is 3.1 105 m and the distance from the slit to the screen is 2.2 m. If the beam of light of wavelength 600 nm passes through
More informationGeometric Optics Converging Lenses and Mirrors Physics Lab IV
Objective Geometric Optics Converging Lenses and Mirrors Physics Lab IV In this set of lab exercises, the basic properties geometric optics concerning converging lenses and mirrors will be explored. The
More information1. You stand two feet away from a plane mirror. How far is it from you to your image? a. 2.0 ft c. 4.0 ft b. 3.0 ft d. 5.0 ft
Lenses and Mirrors 1. You stand two feet away from a plane mirror. How far is it from you to your image? a. 2.0 ft c. 4.0 ft b. 3.0 ft d. 5.0 ft 2. Which of the following best describes the image from
More informationRevision problem. Chapter 18 problem 37 page 612. Suppose you point a pinhole camera at a 15m tall tree that is 75m away.
Revision problem Chapter 18 problem 37 page 612 Suppose you point a pinhole camera at a 15m tall tree that is 75m away. 1 Optical Instruments Thin lens equation Refractive power Cameras The human eye Combining
More informationChapter 36  Lenses. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University
Chapter 36  Lenses A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 2007 Objectives: After completing this module, you should be able to: Determine
More information2) A convex lens is known as a diverging lens and a concave lens is known as a converging lens. Answer: FALSE Diff: 1 Var: 1 Page Ref: Sec.
Physics for Scientists and Engineers, 4e (Giancoli) Chapter 33 Lenses and Optical Instruments 33.1 Conceptual Questions 1) State how to draw the three rays for finding the image position due to a thin
More informationReview for Test 3. Polarized light. Action of a Polarizer. Polarized light. Light Intensity after a Polarizer. Review for Test 3.
Review for Test 3 Polarized light No equation provided! Polarized light In linearly polarized light, the electric field vectors all lie in one single direction. Action of a Polarizer Transmission axis
More informationStudy Guide for Exam on Light
Name: Class: Date: Study Guide for Exam on Light Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which portion of the electromagnetic spectrum is used
More informationGEOMETRICAL OPTICS. Lens Prism Mirror
GEOMETRICAL OPTICS Geometrical optics is the treatment of the passage of light through lenses, prisms, etc. by representing the light as rays. A light ray from a source goes in a straight line through
More informationLecture Notes for Chapter 34: Images
Lecture Notes for hapter 4: Images Disclaimer: These notes are not meant to replace the textbook. Please report any inaccuracies to the professor.. Spherical Reflecting Surfaces Bad News: This subject
More informationThin Lenses Drawing Ray Diagrams
Drawing Ray Diagrams Fig. 1a Fig. 1b In this activity we explore how light refracts as it passes through a thin lens. Eyeglasses have been in use since the 13 th century. In 1610 Galileo used two lenses
More informationPHYS 222 Spring 2012 Final Exam. Closed books, notes, etc. No electronic device except a calculator.
PHYS 222 Spring 2012 Final Exam Closed books, notes, etc. No electronic device except a calculator. NAME: (all questions with equal weight) 1. If the distance between two point charges is tripled, the
More informationRESOLVING POWER OF A READING TELESCOPE
96 Lab Experiments Experiment255 RESOLVING POWER OF A READING TELESCOPE S Dr Jeethendra Kumar P K KamalJeeth Instrumentation & Service Unit, No60, TATA Nagar, Bangalore560 092, INDIA. Email:jeeth_kjisu@rediffmail.com
More informationwaves rays Consider rays of light from an object being reflected by a plane mirror (the rays are diverging): mirror object
PHYS1000 Optics 1 Optics Light and its interaction with lenses and mirrors. We assume that we can ignore the wave properties of light. waves rays We represent the light as rays, and ignore diffraction.
More informationEXPERIMENT 6 OPTICS: FOCAL LENGTH OF A LENS
EXPERIMENT 6 OPTICS: FOCAL LENGTH OF A LENS The following website should be accessed before coming to class. Text reference: pp189196 Optics Bench a) For convenience of discussion we assume that the light
More informationChapter 23. The Reflection of Light: Mirrors
Chapter 23 The Reflection of Light: Mirrors Wave Fronts and Rays Defining wave fronts and rays. Consider a sound wave since it is easier to visualize. Shown is a hemispherical view of a sound wave emitted
More informationAnswer: b. Answer: a. Answer: d
Practice Test IV Name 1) In a single slit diffraction experiment, the width of the slit is 3.1 105 m and the distance from the slit to the screen is 2.2 m. If the beam of light of wavelength 600 nm passes
More informationPhysics 41 Chapter 38 HW Key
Physics 41 Chapter 38 HW Key 1. Helium neon laser light (63..8 nm) is sent through a 0.300mmwide single slit. What is the width of the central imum on a screen 1.00 m from the slit? 7 6.38 10 sin θ.11
More informationPhysics 10. Lecture 29A. "There are two ways of spreading light: to be the candle or the mirror that reflects it." Edith Wharton
Physics 10 Lecture 29A "There are two ways of spreading light: to be the candle or the mirror that reflects it." Edith Wharton Converging Lenses What if we wanted to use refraction to converge parallel
More informationInterference. Physics 102 Workshop #3. General Instructions
Interference Physics 102 Workshop #3 Name: Lab Partner(s): Instructor: Time of Workshop: General Instructions Workshop exercises are to be carried out in groups of three. One report per group is due by
More informationChapter 32. OPTICAL IMAGES 32.1 Mirrors
Chapter 32 OPTICAL IMAGES 32.1 Mirror The point P i called the image or the virtual image of P (light doe not emanate from it) The leftright reveral in the mirror i alo called the depth inverion (the
More informationPROPERTIES OF THIN LENSES. Paraxialray Equations
PROPERTIES OF THIN LENSES Object: To measure the focal length of lenses, to verify the thin lens equation and to observe the more common aberrations associated with lenses. Apparatus: PASCO Basic Optical
More informationLenses. Types of Lenses (The word lens is derived from the Latin word lenticula, which means lentil. A lens is in the shape of a lentil.
Lenses Notes_10_ SNC2DE_0910 Types of Lenses (The word lens is derived from the Latin word lenticula, which means lentil. A lens is in the shape of a lentil. ) Most lenses are made of transparent glass
More information15 Imaging ESSENTIAL IDEAS. How we see images. Option C. Understanding the human eye
Option C 15 Imaging ESSENTIAL IDEAS The progress of a wave can be modelled using the ray or the wavefront. The change in wave speed when moving between media changes the shape of the wave. Optical microscopes
More informationLesson 26: Reflection & Mirror Diagrams
Lesson 26: Reflection & Mirror Diagrams The Law of Reflection There is nothing really mysterious about reflection, but some people try to make it more difficult than it really is. All EMR will reflect
More informationLaw of Reflection. The angle of incidence (i) is equal to the angle of reflection (r)
Light GCSE Physics Reflection Law of Reflection The angle of incidence (i) is equal to the angle of reflection (r) Note: Both angles are measured with respect to the normal. This is a construction line
More informationPhysics 25 Exam 3 November 3, 2009
1. A long, straight wire carries a current I. If the magnetic field at a distance d from the wire has magnitude B, what would be the the magnitude of the magnetic field at a distance d/3 from the wire,
More informationC) D) As object AB is moved from its present position toward the left, the size of the image produced A) decreases B) increases C) remains the same
1. For a plane mirror, compared to the object distance, the image distance is always A) less B) greater C) the same 2. Which graph best represents the relationship between image distance (di) and object
More informationGeometric Optics Physics 118/198/212. Geometric Optics
Background Geometric Optics This experiment deals with image formation with lenses. We will use what are referred to as thin lenses. Thin lenses are ordinary lenses like eyeglasses and magnifiers, but
More informationDiffraction and Young s Single Slit Experiment
Diffraction and Young s Single Slit Experiment Developers AB Overby Objectives Preparation Background The objectives of this experiment are to observe Fraunhofer, or farfield, diffraction through a single
More informationLesson 29: Lenses. Double Concave. Double Convex. Planoconcave. Planoconvex. Convex meniscus. Concave meniscus
Lesson 29: Lenses Remembering the basics of mirrors puts you half ways towards fully understanding lenses as well. The same sort of rules apply, just with a few modifications. Keep in mind that for an
More informationLecture 12: Fraunhofer diffraction by a single slit
Lecture 12: Fraunhofer diffraction y a single slit Lecture aims to explain: 1. Diffraction prolem asics (reminder) 2. Calculation of the diffraction integral for a long slit 3. Diffraction pattern produced
More informationDiffraction of Laser Light
Diffraction of Laser Light No Prelab Introduction The laser is a unique light source because its light is coherent and monochromatic. Coherent light is made up of waves, which are all in phase. Monochromatic
More informationDetermination of Focal Length of A Converging Lens and Mirror
Physics 41 Lab 5 Determination of Focal Length of A Converging Lens and Mirror Objective: Apply the thinlens equation and the mirror equation to determine the focal length of a converging (biconvex)
More informationLight and its effects
Light and its effects Light and the speed of light Shadows Shadow films Pinhole camera (1) Pinhole camera (2) Reflection of light Image in a plane mirror An image in a plane mirror is: (i) the same size
More information1051232 Imaging Systems Laboratory II. Laboratory 4: Basic Lens Design in OSLO April 2 & 4, 2002
05232 Imaging Systems Laboratory II Laboratory 4: Basic Lens Design in OSLO April 2 & 4, 2002 Abstract: For designing the optics of an imaging system, one of the main types of tools used today is optical
More informationHOMEWORK 4 with Solutions
Winter 996 HOMEWORK 4 with Solutions. ind the image of the object for the single concave mirror system shown in ig. (see next pages for worksheets) by: (a) measuring the radius R and calculating the focal
More informationProcedure: Geometrical Optics. Theory Refer to your Lab Manual, pages 291 294. Equipment Needed
Theory Refer to your Lab Manual, pages 291 294. Geometrical Optics Equipment Needed Light Source Ray Table and Base Threesurface Mirror Convex Lens Ruler Optics Bench Cylindrical Lens Concave Lens Rhombus
More informationPhysics 202 Problems  Week 8 Worked Problems Chapter 25: 7, 23, 36, 62, 72
Physics 202 Problems  Week 8 Worked Problems Chapter 25: 7, 23, 36, 62, 72 Problem 25.7) A light beam traveling in the negative z direction has a magnetic field B = (2.32 10 9 T )ˆx + ( 4.02 10 9 T )ŷ
More informationEXPERIMENT O6. Michelson Interferometer. Abstract. References. PreLab
EXPERIMENT O6 Michelson Interferometer Abstract A Michelson interferometer, constructed by the student, is used to measure the wavelength of HeNe laser light and the index of refraction of a flat transparent
More informationOPTICAL IMAGES DUE TO LENSES AND MIRRORS *
1 OPTICAL IMAGES DUE TO LENSES AND MIRRORS * Carl E. Mungan U.S. Naval Academy, Annapolis, MD ABSTRACT The properties of real and virtual images formed by lenses and mirrors are reviewed. Key ideas are
More informationLecture 17. Image formation Ray tracing Calculation. Lenses Convex Concave. Mirrors Convex Concave. Optical instruments
Lecture 17. Image formation Ray tracing Calculation Lenses Convex Concave Mirrors Convex Concave Optical instruments Image formation Laws of refraction and reflection can be used to explain how lenses
More informationLIGHT REFLECTION AND REFRACTION
QUESTION BANK IN SCIENCE CLASSX (TERMII) 10 LIGHT REFLECTION AND REFRACTION CONCEPTS To revise the laws of reflection at plane surface and the characteristics of image formed as well as the uses of reflection
More informationBasic Optics System OS8515C
40 50 30 60 20 70 10 80 0 90 80 10 20 70 T 30 60 40 50 50 40 60 30 C 70 20 80 10 90 90 0 80 10 70 20 60 50 40 30 Instruction Manual with Experiment Guide and Teachers Notes 01209900B Basic Optics System
More informationYour Comments. Also, that 30/60/90 triangle prism question on the last homework...holy geometry batman!
Your Comments Also, that 30/60/90 triangle prism question on the last homework...holy geometry batman! o thats why I'm always up side down when I look at the inside o my cereal spoon, regardless o how
More informationPage 1 Class 10 th Physics LIGHT REFLECTION AND REFRACTION
Page 1 LIGHT Light is a form of energy, which induces the sensation of vision in our eyes and makes us able to see various things present in our surrounding. UNITS OF LIGHT Any object which has an ability
More informationPhysics 116. Nov 4, 2011. Session 22 Review: ray optics. R. J. Wilkes Email: ph116@u.washington.edu
Physics 116 Session 22 Review: ray optics Nov 4, 2011 R. J. Wilkes Email: ph116@u.washington.edu ! Exam 2 is Monday!! All multiple choice, similar to HW problems, same format as Exam 1!!! Announcements
More informationPhysics 117.3 Tutorial #1 January 14 to 25, 2013
Physics 117.3 Tutorial #1 January 14 to 25, 2013 Rm 130 Physics 8.79. The location of a person s centre of gravity can be determined using the arrangement shown in the figure. A light plank rests on two
More information4.1: Angles and Radian Measure
4.1: Angles and Radian Measure An angle is formed by two rays that have a common endpoint. One ray is called the initial side and the other is called the terminal side. The endpoint that they share is
More informationInterference and Diffraction
Chapter 14 nterference and Diffraction 14.1 Superposition of Waves... 1414. Young s DoubleSlit Experiment... 144 Example 14.1: DoubleSlit Experiment... 147 14.3 ntensity Distribution... 148 Example
More informationChapter 17: Light and Image Formation
Chapter 17: Light and Image Formation 1. When light enters a medium with a higher index of refraction it is A. absorbed. B. bent away from the normal. C. bent towards from the normal. D. continues in the
More informationExperiment 3 Lenses and Images
Experiment 3 Lenses and Images Who shall teach thee, unless it be thine own eyes? Euripides (480?406? BC) OBJECTIVES To examine the nature and location of images formed by es. THEORY Lenses are frequently
More informationImage Formation Principle
Image Formation Principle Michael Biddle Robert Dawson Department of Physics and Astronomy The University of Georgia, Athens, Georgia 30602 (Dated: December 8, 2015) The aim of this project was to demonstrate
More information9/16 Optics 1 /11 GEOMETRIC OPTICS
9/6 Optics / GEOMETRIC OPTICS PURPOSE: To review the basics of geometric optics and to observe the function of some simple and compound optical devices. APPARATUS: Optical bench, lenses, mirror, target
More informationOptics and Geometry. with Applications to Photography Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles November 15, 2004
Optics and Geometry with Applications to Photography Tom Davis tomrdavis@earthlink.net http://www.geometer.org/mathcircles November 15, 2004 1 Useful approximations This paper can be classified as applied
More informationChapter 22: Mirrors and Lenses
Chapter 22: Mirrors and Lenses How do you see sunspots? When you look in a mirror, where is the face you see? What is a burning glass? Make sure you know how to:. Apply the properties of similar triangles;
More informationChapter 27 Optical Instruments. 27.1 The Human Eye and the Camera 27.2 Lenses in Combination and Corrective Optics 27.3 The Magnifying Glass
Chapter 27 Optical Instruments 27.1 The Human Eye and the Camera 27.2 Lenses in Combination and Corrective Optics 27.3 The Magnifying Glass Figure 27 1 Basic elements of the human eye! Light enters the
More informationWAVELENGTH OF LIGHT  DIFFRACTION GRATING
PURPOSE In this experiment we will use the diffraction grating and the spectrometer to measure wavelengths in the mercury spectrum. THEORY A diffraction grating is essentially a series of parallel equidistant
More informationWaves and Light Extra Study Questions
Waves and Light Extra Study Questions Short Answer 1. Determine the frequency for each of the following. (a) A bouncing spring completes 10 vibrations in 7.6 s. (b) An atom vibrates 2.5 10 10 times in
More informationPHY208FALL2008. Week2HW. Introduction to TwoSource Interference. Due at 11:59pm on Friday, September 12, View Grading Details [ Print ]
Assignment Display Mode: View Printable Answers PHY208FALL2008 Week2HW Due at 11:59pm on Friday September 12 2008 View Grading Details [ Print ] The following three problems concern interference from two
More informationReflection and Refraction
Equipment Reflection and Refraction Acrylic block set, planeconcaveconvex universal mirror, cork board, cork board stand, pins, flashlight, protractor, ruler, mirror worksheet, rectangular block worksheet,
More informationINTERFERENCE OBJECTIVES PRELECTURE. Aims
53 L4 INTERFERENCE Aims OBJECTIVES When you have finished this chapter you should understand how the wave model of light can be used to explain the phenomenon of interference. You should be able to describe
More information19  RAY OPTICS Page 1 ( Answers at the end of all questions )
19  RAY OPTICS Page 1 1 ) A ish looking up through the water sees the outside world contained in a circular horizon. I the reractive index o water is 4 / 3 and the ish is 1 cm below the surace, the radius
More informationChapter 2 Laser Diode Beam Propagation Basics
Chapter 2 Laser Diode Beam Propagation Basics Abstract Laser diode beam propagation characteristics, the collimating and focusing behaviors and the M 2 factor are discussed using equations and graphs.
More informationIt bends away from the normal, like this. So the angle of refraction, r is greater than the angle of incidence, i.
Physics 1403 Lenses It s party time, boys and girls, because today we wrap up our study of physics. We ll get this party started in a bit, but first, you have some more to learn about refracted light.
More informationQuestion 2: The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Question 1: Define the principal focus of a concave mirror. ANS: Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting
More informationFraunhofer Diffraction
Physics 334 Spring 1 Purpose Fraunhofer Diffraction The experiment will test the theory of Fraunhofer diffraction at a single slit by comparing a careful measurement of the angular dependence of intensity
More informationIt s shape. Sound Beams foundation for a basic understanding. Anatomy of the beam. Focus. Near Zone. Focal Length. Chapter 10
It s shape Sound Beams foundation for a basic understanding It begins the size or diameter of the transducer gradually converges to a narrower point then begins to diverge Chapter 10 Anatomy of the beam
More informationRefraction of Light at a Plane Surface. Object: To study the refraction of light from water into air, at a plane surface.
Refraction of Light at a Plane Surface Object: To study the refraction of light from water into air, at a plane surface. Apparatus: Refraction tank, 6.3 V power supply. Theory: The travel of light waves
More informationSize Of the Image Nature Of the Image At Infinity At the Focus Highly Diminished, Point Real and Inverted
CHAPTER10 LIGHT REFLECTION AND REFRACTION Light rays; are; electromagnetic in nature, and do not need material medium for Propagation Speed of light in vacuum in 3*10 8 m/s When a light ray falls on a
More informationLab 9. Optics. 9.1 Introduction
Lab 9 Name: Optics 9.1 Introduction Unlike other scientists, astronomers are far away from the objects they want to examine. Therefore astronomers learn everything about an object by studying the light
More informationM01/430/H(3) Name PHYSICS HIGHER LEVEL PAPER 3. Number. Wednesday 16 May 2001 (morning) 1 hour 15 minutes INSTRUCTIONS TO CANDIDATES
INTERNATIONAL BACCALAUREATE BACCALAURÉAT INTERNATIONAL BACHILLERATO INTERNACIONAL M01/430/H(3) PHYSICS HIGHER LEVEL PAPER 3 Wednesday 16 May 2001 (morning) Name Number 1 hour 15 minutes INSTRUCTIONS TO
More informationPhysics, Chapter 38: Mirrors and Lenses
University of Nebraska  Lincoln DigitalCommons@University of Nebraska  Lincoln Robert Katz Publications Research Papers in Physics and Astronomy 111958 Physics, Chapter 38: Mirrors and Lenses Henry
More informationPHYSICS PAPER 1 (THEORY)
PHYSICS PAPER 1 (THEORY) (Three hours) (Candidates are allowed additional 15 minutes for only reading the paper. They must NOT start writing during this time.) 
More informationPreCalculus Review Problems Solutions
MATH 1110 (Lecture 00) August 0, 01 1 Algebra and Geometry PreCalculus Review Problems Solutions Problem 1. Give equations for the following lines in both pointslope and slopeintercept form. (a) The
More informationPhysical Science Study Guide Unit 7 Wave properties and behaviors, electromagnetic spectrum, Doppler Effect
Objectives: PS7.1 Physical Science Study Guide Unit 7 Wave properties and behaviors, electromagnetic spectrum, Doppler Effect Illustrate ways that the energy of waves is transferred by interaction with
More informationUsing light scattering method to find The surface tension of water
Experiment (8) Using light scattering method to find The surface tension of water The aim of work: The goals of this experiment are to confirm the relationship between angular frequency and wave vector
More informationTHE COMPOUND MICROSCOPE
THE COMPOUND MICROSCOPE In microbiology, the microscope plays an important role in allowing us to see tiny objects that are normally invisible to the naked eye. It is essential for students to learn how
More informationPhysics 30 Worksheet # 14: Michelson Experiment
Physics 30 Worksheet # 14: Michelson Experiment 1. The speed of light found by a Michelson experiment was found to be 2.90 x 10 8 m/s. If the two hills were 20.0 km apart, what was the frequency of the
More informationEighth Grade Electromagnetic Radiation and Light Assessment
Eighth Grade Electromagnetic Radiation and Light Assessment 1a. Light waves are the only waves that can travel through. a. space b. solids 1b. Electromagnetic waves, such as light, are the only kind of
More informationDiffraction of a Circular Aperture
Diffraction of a Circular Aperture Diffraction can be understood by considering the wave nature of light. Huygen's principle, illustrated in the image below, states that each point on a propagating wavefront
More informationCHAPTER 4 OPTICAL ABERRATIONS
1 CHAPTER 4 OPTICAL ABERRATIONS 4.1 Introduction We have hitherto made the assumption that a lens or a curved mirror is able to form a point image of a point object. This may be approximately true if the
More informationOptics Formulas. Light Intensity
LENS SELECTION GUIDE 494 Optics Light RightHand Rule Light is a transverse electromagnetic wave. The electric E and magnetic M fields are perpendicular to each other and to the propagation vector k, as
More informationBasic Physical Optics
F UNDAMENTALS OF PHOTONICS Module 1.4 Basic Physical Optics Leno S. Pedrotti CORD Waco, Texas In Module 13, Basic Geometrical Optics, we made use of light rays to demonstrate reflection and refraction
More information1 Basic Optics (1.2) Since. ε 0 = 8.854 10 12 C 2 N 1 m 2 and μ 0 = 4π 10 7 Ns 2 C 2 (1.3) Krishna Thyagarajan and Ajoy Ghatak. 1.
1 1 Basic Optics Krishna Thyagarajan and Ajoy Ghatak 1.1 Introduction This chapter on optics provides the reader with the basic understanding of light rays and light waves, image formation and aberrations,
More informationMeasuring the Point Spread Function of a Fluorescence Microscope
Frederick National Laboratory Measuring the Point Spread Function of a Fluorescence Microscope Stephen J Lockett, PhD Principal Scientist, Optical Microscopy and Analysis Laboratory Frederick National
More informationScience In Action 8 Unit C  Light and Optical Systems. 1.1 The Challenge of light
1.1 The Challenge of light 1. Pythagoras' thoughts about light were proven wrong because it was impossible to see A. the light beams B. dark objects C. in the dark D. shiny objects 2. Sir Isaac Newton
More informationRAY OPTICS II 7.1 INTRODUCTION
7 RAY OPTICS II 7.1 INTRODUCTION This chapter presents a discussion of more complicated issues in ray optics that builds on and extends the ideas presented in the last chapter (which you must read first!)
More informationGRID AND PRISM SPECTROMETERS
FYSA230/2 GRID AND PRISM SPECTROMETERS 1. Introduction Electromagnetic radiation (e.g. visible light) experiences reflection, refraction, interference and diffraction phenomena when entering and passing
More information3.5.4.2 One example: Michelson interferometer
3.5.4.2 One example: Michelson interferometer mirror 1 mirror 2 light source 1 2 3 beam splitter 4 object (n object ) interference pattern we either observe fringes of same thickness (parallel light) or
More informationThe light. Light (normally spreads out straight... ... and into all directions. Refraction of light
The light Light (normally spreads out straight...... and into all directions. Refraction of light But when a light ray passes from air into glas or water (or another transparent medium), it gets refracted
More informationCrystal Optics of Visible Light
Crystal Optics of Visible Light This can be a very helpful aspect of minerals in understanding the petrographic history of a rock. The manner by which light is transferred through a mineral is a means
More information