WAVES

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WAVES

• Optics

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WAVE BEHAVIOR 3 DIFFRACTION

• Diffraction is the bending of a wave AROUND a
  barrier
• Diffracted waves can interfere and cause
  diffraction patterns

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DOUBLE SLIT DIFFRACTION

• n? d sinT
• n bright band number
• ? wavelength (m)
• d space between slits (m)
• T angle defined by central band, slit, and band n

• This also works for diffraction gratings
  consisting of many, many slits that allow the
  light to pass through. Each slit acts as a
  separate light source

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SINGLE SLIT DIFFRACTION

• n ? s sin T
• n dark band number
• ? wavelength (m)
• s slit width (m)
• T angle defined by central band, slit, and dark
  band

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• Light of wavelength 360 nm is passed through a
  diffraction grating that has 10,000 slits per cm.
  If the screen is 2.0 m from the grating, how far
  from the central bright band is the first order
  bright band?

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• Light of wavelength 560 nm is passed through two
  slits. It is found that, on a screen 1.0 m from
  the slits, a bright spot is formed at x 0, and
  another is formed at x 0.03 m. What is the
  spacing between the slits?

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REFLECTION

• Reflected sound can be heard as an echo
• Light waves can be drawn as rays to diagram
  light reflected off mirrors

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REFLECTION AND PLANE MIRRORS

• Law of Reflection

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MIRRORS

• Mirrors can be
• Plane (flat)
• Spherical
• Convex reflective side curves outward
• Concave reflective side curves inward

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OPTICAL IMAGES

• Nature
• Real (converging rays)
• Virtual (diverging rays)
• Orientation
• Upright
• Inverted
• Size
• True
• Enlarged
• Reduced

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MIRRORS AND RAY TRACING

• Ray tracing is a method of constructing an image
  using the model of light as a ray
• We use ray tracing to construct optical images
  produced by mirrors and lenses
• Ray tracing lets us describe what happens to the
  light as it interacts with a medium

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RAY TRACING PLANE MIRRORS

• Use at least two rays to construct the image

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PROBLEM 4

• Standing 2.0 m in front of a small vertical
  mirror, you see the reflection of your belt
  buckle, which is 0.70 m below your eyes.
• What is the vertical location of the mirror
  relative to the level of your eyes?
• If you move backward until you are 6.0 m from the
  mirror, will you still see the buckle, or will
  you see a point on your body that is above or
  below the buckle? Explain.

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SOLUTION
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SPHERICAL MIRRORS

• Concave

• Convex

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PARTS OF ASPHERICAL CONCAVE MIRROR
-

Vertex
Center
Principle axis
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PARTS OF A SPHERICAL CONCAVE MIRROR

• The focal length is half the radius of curvature
• R 2f
• The focal length is positive for a concave mirror
  because it is on the shiny side
• Rays parallel to the optical axis all pas through
  the focus

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RAY TRACING FOR CONCAVE MIRRORS

• You must draw at least TWO of the three principal
  rays to construct an image
• The p-ray parallel to the principal axis, then
  reflects through the focus
• The f-ray travels through the focus then
  reflects back parallel to the principal axis
• The c-ray travels through the center, then
  reflects back through the center

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CONCAVE MIRRORS

• Diagram the following images
• Object outside the center of curvature
• Object at the center of curvature
• Object between center of curvature and focus
• Object at the focal point
• Object inside the focus

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SPHERICAL CONCAVE MIRROR(OBJECT OUTSIDE CENTER)
c
p
Real, Inverted, Reduced Image
f
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SPHERICAL CONCAVE MIRROR(OBJECT AT CENTER)
Real, Inverted, True Image
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SPHERICAL CONCAVE MIRROR(OBJECT BETWEEN CENTER
AND FOCUS)
Real, Inverted, EnlargedImage
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SPHERICAL CONCAVE MIRROR(OBJECT AT FOCUS)
No image
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SPHERICAL CONCAVE MIRROR(OBJECT INSIDE FOCUS)
Virtual, Upright, Enlarged Image
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MIRROR EQUATION 1 MIRROR EQUATION 2

• 1/si 1/s0 1/f
• si image distance
• s0 object distance
• f focal length

• M hi/h0 -si/s0
• si image distance
• s0 object distance
• hi image height
• h0 object height
• M magnification

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SIGN CONVENTIONS

• Focal length (f)
• Positive for concave mirrors
• Negative for convex mirrors
• Magnification (M)
• Positive for upright images
• Negative for inverted images
• Enlarged M gt 1
• Reduced M lt 1
• Image Distance
• si is positive for real images
• si is negative for virtual images

• Practice A spherical concave mirror, focal
  length 20 cm, has a 5-cm high object placed 30 cm
  from it
• Draw a ray diagram and construct the image
• Use mirror equations to calculate the position,
  magnification, and size of the image
• Name the image

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• Solution A spherical concave mirror, focal
  length 20 cm, has a 5-cm high object placed 30 cm
  from it.
• Calculate the position, magnification, and size
  of the image.
• Name the image

real, inverted, enlarged.
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PARTS OF A SPHERICAL CONVEX MIRROR

• The focal length is half the radius of curvature
  and both are on the dark side of the mirror
• The focal length is negative

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RAY TRACING

• Construct the image for an object located outside
  a spherical convex mirror
• Name the image

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PRACTICE 5

• A spherical concave mirror, focal length 15 cm,
  has a 4-cm high object placed 10 cm from it
• Draw a ray diagram and construct the image
• Use the mirror equations to calculate position,
  magnification, and size of image
• Name the image

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• Problem A spherical convex mirror, focal length
  15 cm, has a 4-cm high object placed 10 cm from
  it.
• a) Use the mirror equations to calculate
• the position of image
• the magnification
• the size of image
• b) Name the image virtual, upright,
  reduced size

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SPHERICAL MIRRORS

• Concave

• Convex

• Image is real when object is outside focus
• Image is virtual when object is inside focus
• Focal length is positive

• Image is ALWAYS virtual
• Focal length is negative

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Real vs Virtual images

• Real
• Formed by converging light rays
• si is positive when calculated with mirror
  equation

• Virtual
• Formed by diverging light rays
• si is negative when calculated with mirror
  equation

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Upright vs Inverted images

• Upright
• Always virtual if formed by one mirror or lens
• hi is positive when calculated with mirror/lens
  equation

• Inverted
• Always real if formed by one mirror or lens
• hi is negative when calculated with mirror/lens
  equation

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WAVE BEHAVIOR 2 REFRACTION

• Refraction occurs when a wave is transmitted from
  one medium to another
• Refracted waves may change speed and wavelength
• Refraction is almost always accompanied by some
  reflection
• Refracted waves do NOT change frequency

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REFRACTION OF LIGHT

• Refraction causes a change in speed of light as
  it moves from one medium to another
• Refraction can cause bending of the light ray at
  the interface between media
• Index of Refraction (n)
• n speed of light in vacuum / speed of light in
  medium
• n c/v

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SNELLS LAW

• n1sinT1 n2sinT2

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SNELLS LAW

• When the index of refraction increases, light
  bends TOWARD the normal
• When the index of refraction decreases, light
  bends AWAY FROM the normal

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PROBLEM 6!

• Light enters an oil from the air at an angle of
  50 with the normal, and the refracted beam makes
  an angle of 33 with the normal
• Draw the situation
• Calculate the index of refraction of the oil
• Calculate the speed of light in the oil

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• Solution Light enters an oil from the air at an
  angle of 50o with the normal, and the refracted
  beam makes an angle of 33o with the normal.
• Draw this situation.
• Calculate the index of refraction of the oil.
• Calculate the speed of light in the oil

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PRISM PROBLEMS (7)

• Light enters a prism as shown, and passes through
  the prism
• a. Complete the path of light through the prism
  and show the angle it will make when it leaves
  the prism
• b. If the refractive index of the glass is 1.55,
  calculate the angle of refraction when it leaves
  the prism
• c. How would the answer to (b) change if the
  prism were immersed in water?

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• Solution Light enters a prism as shown, and
  passes through the prism.
• Complete the path of the light through the prism,
  and show the angle it will make when it leaves
  the prism.
• If the refractive index of the glass is 1.55,
  calculate the angle of refraction when it leaves
  the prism.
• How would the answer to b) change if the prism
  were immersed in water?

q2
c) In water, the angle of the light as it leaves
the glass would be smaller, since the indices of
refraction would be more similar and there would
be less bending.
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PRISM PROBLEMS (8)

• Light enters a prism made of air from glass
• Complete the path of the light through the prism,
  and show the angle it will make when it leaves
  the prism
• If the refractive index of the glass is 1.55,
  calculate the angle of refraction when it leaves
  the prism

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• Problem Light enters a prism made of air from
  glass.
• Complete the path of the light through the prism,
  and show the angle it will make when it leaves
  the prism.
• If the refractive index of the glass is 1.55,
  calculate the angle of refraction when it leaves
  the prism.

30o
q2
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CRITICAL ANGLE OF INCIDENCE

• The smallest angle of incidence for which light
  cannot leave a medium is called the critical
  angle of incidence
• If light passes into a medium with a greater
  refractive index than the original medium, it
  bends away from the normal and the angle of
  refraction is greater than the angle of incidence
• If the angle of refraction is 90, the light
  cannot leave the medium and no refraction occurs
• We call this TOTAL INTERNAL REFLECTION

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TOTAL INTERNAL REFLECTION

• Calculating the Critical Angle

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PRACTICE NO. 9

• What is the critical angle of incidence for a
  gemstone with refractive index 2.45 if it is in
  air?
• If you immerse it in water (refractive index
  1.33), what does this do to the critical angle of
  incidence?

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• Solution What is the critical angle of incidence
  for a gemstone with refractive index 2.45 if it
  is in air?
• If you immerse the gemstone in water (refractive
  index 1.33), what does this do to the critical
  angle of incidence?
• It increases the critical angle of incidence
  because there is less difference in the
  refractive indices.

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LENSES REFRACT LIGHT

• Converging (Convex)

• Diverging (Concave)

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LENS RAY TRACING

• Ray tracing is used for lenses also. Use the
  same principal rays used with mirrors. You must
  draw TWO of the three
• the p-ray parallel to the principal axis,
  refracts through the focus
• the f-ray travels through the focus, then
  refracts parallel to the principal axis
• the c-ray travels through the center and
  continues without bending
• Use the same equations we used for mirrors

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CONVERGING LENSES
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CONSTRUCT THE FOLLOWING IMAGES

• Object located outside 2F for a converging lens
• Object located at 2F
• Object located between F and 2F
• Object at the focus
• Object inside the focus

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FOR CONVERGING LENSES

• f is positive
• so is positive
• si is positive for real images and negative for
  virtual images
• M is negative for real images and positive for
  virtual images
• hi is negative for real images and positive for
  virtual images

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DIVERGING LENSES

• Construct an image for an object located in front
  of a diverging lens

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FOR DIVERGING LENSES

• f is negative
• so is positive
• si is negative
• M is positive and lt 1
• hi is positive and lt ho