Lecture 14 Images Chapter 34
Download
1 / 49

Lecture 14 Images Chapter 34 - PowerPoint PPT Presentation


  • 114 Views
  • Uploaded on

Lecture 14 Images Chapter 34. Geometrical Optics Fermats Principle Law of reflection Law of Refraction Plane Mirrors and Spherical Mirrors Spherical refracting Surfaces Thin Lenses Optical Instruments Magnifying Glass, Microscope, Refracting telescope Polling Questions.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Lecture 14 Images Chapter 34' - chelsey


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
Slide1 l.jpg

Lecture 14 Images Chapter 34

  • Geometrical Optics

  • Fermats Principle

    • Law of reflection

    • Law of Refraction

  • Plane Mirrors and Spherical Mirrors

  • Spherical refracting Surfaces

  • Thin Lenses

  • Optical Instruments

    • Magnifying Glass, Microscope, Refracting telescope

  • Polling Questions


  • Slide2 l.jpg

    Left over from Lecture 13 and Chapter 33

    • Show electric and magnetic fields are out of phase in demo.

    • Reflection and Refraction

      • Red or green laser through smoky block

      • Show maximum bending angle

    • Total internal reflection

    • 1) Show it with block

    • Dispersion through prism

    • Brewsters Law

      • Show using polaroid paper



    Slide4 l.jpg

    Dispersion: Different wavelengths have different velocities and therefore different indices of refraction. This leads to differentrefractive angles for different wavelengths. Thus the light is dispersed.The frequency does not change when n changes.


    Snells law l.jpg

    n and therefore different indices of refraction. This leads to different2=1.509

    Snells Law

    Red

    θ1

    n1=1.000

    θ2


    Snell s law at work l.jpg
    Snell’s Law at Work and therefore different indices of refraction. This leads to different


    Fiber cable l.jpg
    Fiber Cable and therefore different indices of refraction. This leads to different



    Slide9 l.jpg

    Equilateral prism Internal reflection

    dispersing sunlight

    late afternoon.

    Sin θrefr= 0.866nλ


    Slide10 l.jpg

    How much light is polarized when reflected from a surface? Internal reflection

    Polarization by Reflection: Brewsters Law


    Slide11 l.jpg

    Fresnel Equations Internal reflection

    The parallel and perpendicular components of the electric vector of the reflected and transmitted

    light wave is plotted below when light strikes a piece of glass. Now the intensity of polarized

    light is proportional to the square of the amplitude of the oscillations of the electric field. So, we

    can express the intensity of the incoming light as I0 = (constant) E 2.

    E


    Slide12 l.jpg

    What causes a Mirage Internal reflection

    eye

    sky

    1.09

    1.09

    1.08

    1.08

    1.07

    1.07

    Index of refraction

    1.06

    Hot road causes gradient in the index of refraction that increases

    as you increase the distance from the road


    Inverse mirage bend l.jpg
    Inverse Mirage Bend Internal reflection


    Slide14 l.jpg

    Geometrical Optics Internal reflection:Study of reflection and refraction of light from surfaces

    The ray approximation states that light travels in straight lines

    until it is reflected or refracted and then travels in straight lines again.

    The wavelength of light must be small compared to the size of

    the objects or else diffractive effects occur.


    Law of reflection l.jpg
    Law of Reflection Internal reflection

    Mirror

    B

    A

    1


    Fermat s principle l.jpg
    Fermat’s Principle Internal reflection

    Using Fermat’s Principle you can prove the

    Reflection law. It states that the path taken

    by light when traveling from one point to

    another is the path that takes the shortest

    time compared to nearby paths.

    JAVA APPLET

    Show Fermat’s principle simulator

    http://www.phys.hawaii.edu/~teb/java/ntnujava/index.html


    Slide17 l.jpg

    B Internal reflection

    2

    A

    1

    Two light rays 1 and 2 taking different paths between points A and B and reflecting off a vertical mirror

    Plane Mirror

    Use calculus - method

    of minimization


    Slide18 l.jpg

    Write down time as a function of y Internal reflection

    and set the derivative to 0.


    Plane mirrors where is the image formed l.jpg

    Mirrors and Lenses Internal reflection

    Plane Mirrors Where is the image formed


    Plane mirrors l.jpg
    Plane mirrors Internal reflection

    Angle of

    incidence

    Virtual side

    Real side

    Normal

    Virtual image

    Angle of reflection

    i = - p

    eye

    Object distance = - image distance

    Image size = Object size


    Slide22 l.jpg

    eye Internal reflection

    object

    2

    1

    3

    Problem: Two plane mirrors make an angle of 90o. How many images are there for an object placed between them?

    mirror

    mirror


    Slide23 l.jpg

    Using the Law of Reflection to make a bank shot Internal reflection

    Assuming no spin

    Assuming an elastic collision

    No cushion deformation

    d

    d

    pocket


    Slide24 l.jpg

    i = - p magnification = 1 Internal reflection

    What happens if we bend the mirror?

    Concave mirror.

    Image gets magnified.

    Field of view is diminished

    Convex mirror.

    Image is reduced.

    Field of view increased.


    Rules for drawing images for mirrors l.jpg
    Rules for drawing images for mirrors Internal reflection

    • Initial parallel ray reflects through focal point.

    • Ray that passes in initially through focal point reflects parallel

    • from mirror

    • Ray reflects from C the radius of curvature of mirror reflects along

    • itself.

    • Ray that reflects from mirror at little point c is reflected

    • symmetrically

    Lateral magnification = ratio of image height/object height



    Spherical refracting surfaces l.jpg
    Spherical refracting surfaces Internal reflection

    Using Snell’s Law and assuming small

    Angles between the rays with the central

    axis, we get the following formula:


    Slide29 l.jpg

    In Figure 34-35, A beam of parallel light rays from a laser is incident on a solid transparent sphere of index of refraction n.

    Fig. 34-35

    (a) If a point image is produced at the back of the sphere, what is the index of refraction of the sphere?

    (b) What index of refraction, if any, will produce a point image at the center of the sphere? Enter 'none' if necessary.

    Chapter 34 Problem 32


    Slide31 l.jpg

    Apply this equation to Thin Lenses is incident on a solid transparent sphere of index of refraction n. where the thickness is small compared to object distance, image distance, and radius of curvature. Neglect thickness.

    Converging lens

    Diverging lens


    Simple lens model l.jpg
    Simple Lens Model is incident on a solid transparent sphere of index of refraction n.


    Slide33 l.jpg

    Thin Lens Equation is incident on a solid transparent sphere of index of refraction n.

    Lensmaker Equation

    Lateral Magnification for a Lens

    What is the sign convention?


    Sign convention l.jpg

    p is incident on a solid transparent sphere of index of refraction n.

    Sign Convention

    Real side - R

    Virtual side - V

    Light

    r1

    r2

    i

    Real object - distance p is pos on V side (Incident rays are diverging)

    Radius of curvature is pos on R side.

    Real image - distance is pos on R side.

    Virtual object - distance is neg on R side. Incident rays are converging)

    Radius of curvature is neg on the V side.

    Virtual image- distance is neg on the V side.


    Rules for drawing rays to locate images from a lens l.jpg
    Rules for drawing rays to locate images from a lens is incident on a solid transparent sphere of index of refraction n.

    1) A ray initially parallel to the central axis will pass through the focal point.

    2) A ray that initially passes through the focal point will emerge from the lens

    parallel to the central axis.

    3) A ray that is directed towards the center of the lens will go straight through the lens undeflected.


    Real image ray diagram for a converging lens l.jpg
    Real image ray diagram for a converging lens is incident on a solid transparent sphere of index of refraction n.


    Virtual image ray diagram for converging lens l.jpg
    Virtual image ray diagram for converging lens is incident on a solid transparent sphere of index of refraction n.


    Virtual image ray diagram for a diverging lens l.jpg
    Virtual image ray diagram for a diverging lens is incident on a solid transparent sphere of index of refraction n.


    Slide40 l.jpg

    Example is incident on a solid transparent sphere of index of refraction n.

    Virtual side

    Real side

    .

    .

    F1

    p

    F2

    Given a lens with a focal length f = 5 cm and object distance p = +10 cm, find the following: i and m. Is the image real or virtual? Upright or inverted? Draw the 3 rays.


    Slide41 l.jpg

    Example is incident on a solid transparent sphere of index of refraction n.

    Virtual side

    Real side

    .

    .

    F1

    p

    F2

    Given a lens with a focal length f = 5 cm and object distance p = +10 cm, find the following: i and m. Is the image real or virtual? Upright or inverted? Draw the 3 rays.

    Image is real,

    inverted.


    Slide42 l.jpg

    . is incident on a solid transparent sphere of index of refraction n.

    .

    r1

    r2

    F1

    p

    F2

    Example

    Given a lens with the properties (lengths in cm) r1 = +30, r2 = -30, p = +10, and n = 1.5, find the following: f, i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays.

    Real side

    Virtual side


    Slide43 l.jpg

    . is incident on a solid transparent sphere of index of refraction n.

    .

    r1

    r2

    F1

    p

    F2

    Example

    Given a lens with the properties (lengths in cm) r1 = +30, r2 = -30, p = +10, and n = 1.5, find the following: f, i and m. Is the image real or virtual? Upright or inverted? Draw 3 rays.

    Real side

    Virtual side

    Image is virtual, upright.


    Slide44 l.jpg

    Lens 1 is incident on a solid transparent sphere of index of refraction n.

    Lens 2

    +20

    -15

    f1

    f2

    f1

    f2

    40

    10

    Example

    A converging lens with a focal length of +20 cm is located 10 cm to the left of a diverging lens having a focal length of -15 cm. If an object is located 40 cm to the left of the converging lens, locate and describe completely the final image formed by the diverging lens. Treat each lens

    Separately.


    Slide45 l.jpg

    Lens 1 is incident on a solid transparent sphere of index of refraction n.

    Lens 2

    30

    40

    +20

    -15

    f1

    f2

    f1

    f2

    40

    10

    Ignoring the diverging lens (lens 2), the image formed by the

    converging lens (lens 1) is located at a distance

    Since m = -i1/p1= - 40/40= - 1 , the image is inverted

    This image now serves as a virtual object for lens 2, with p2 = - (40 cm - 10 cm) = - 30 cm.


    Slide46 l.jpg

    Lens 1 is incident on a solid transparent sphere of index of refraction n.

    Lens 2

    30

    40

    +20

    -15

    f1

    f2

    f1

    f2

    40

    10

    Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It is virtual (since i2 < 0).

    The magnification is m = (-i1/p1) x (-i2/p2) = (-40/40)x(30/-30) =+1, so the image

    has the same size orientation as the object.


    Optical instruments l.jpg
    Optical Instruments is incident on a solid transparent sphere of index of refraction n.

    Magnifying lens

    Compound microscope

    Refracting telescope

    Galileo - converging + diverging lens

    Keplerian - converging + converging lens

    Reflecting Telescope


    Chapter 34 problem 91 l.jpg
    Chapter 34 Problem 91 is incident on a solid transparent sphere of index of refraction n.

    Figure 34-43a shows the basic structure of a human eye. Light refracts into the eye through the cornea and is then further redirected by a lens whose shape (and thus ability to focus the light) is controlled by muscles. We can treat the cornea and eye lens as a single effective thin lens (Figure 34-43b). A "normal" eye can focus parallel light rays from a distant object O to a point on the retina at the back of the eye, where processing of the visual information begins. As an object is brought close to the eye, however, the muscles must change the shape of the lens so that rays form an inverted real image on the retina (Figure 34-43c).


    Slide49 l.jpg

    (a) Suppose that for the parallel rays of Figure 34-43a and Figure 34-43b, the focal length f of the effective thin lens of the eye is 2.52 cm. For an object at distance p = 48.0 cm, what focal length f' of the effective lens is required for the object to be seen clearly?

    (b) Must the eye muscles increase or decrease the radii of curvature of the eye lens to give focal length f'?


    ad