LIGHT DEFINITION, light properties etc. the snels law

1. LIGHT
·
Prepared by:
Percy Brenda Q. Modelo
MAEDSci1

2. Learning Objectives
● Describe the nature of light.
● Narrate Maxwell’s line of reasoning in linking EM to light.
● Apply law of reflection and Snell’s law in problem solving.
● Describe different types of mirrors and lenses.
● Relate properties of mirrors and lenses to image and object
distance and sizes.
● Solve problems involving mirror and lens’ equation.

3. 01.
NATURE OF
LIGHT
02.
PROPERTIES
OF LIGHT
03.
SNELL’S
LAW
04.
MIRRORS AND
LENSES

4. NATURE OF
LIGHT
01.

10. PROPERTIES
OF LIGHT
02.

17. REFLECTION
LAWS OF
REFLECTION
The angle of incidence equals
the angle of reflection
The incident ray, reflected ray
and the normal lie on the same
plane

37. SNELL’S LAW
03.

42. MIRRORS
AND LENSES
04.

43. Real Image –
a. Image is made from “real” light
rays that converge at a real focal
point, so the image is REAL
b. Can be projected onto a screen
because light passes through the
point where the image appears
MIRRORS AND LENSES

44. Real Image –
a. Image is made from “real” light
rays that converge at a real focal
point, so the image is REAL
b. Can be projected onto a screen
because light passes through the
point where the image appears
MIRRORS AND LENSES

45. Virtual Image–
a. “Not Real” because it cannot be
projected
b. Image only seems to be there!
MIRRORS AND LENSES

46. Real and Virtual Images:
A real image is one formed by the actual convergence
of light rays in a spot. Real images can be viewed by
placing a screen in the place where they form. By contrast,
virtual images are illusions made by the curved optic
device that trick the eye into thinking that light converged in
a place where it didn't converge. Virtual objects are only
viewed by looking into the lens or mirror forming the image.
MIRRORS AND LENSES

47. Virtual Images in Plane Mirrors
If light energy doesn't flow from the image, the image
is "virtual".
Rays seem to come from behind
the mirror, but, of course, they
don't. It is virtually as if the rays
were coming from behind the
mirror.
"Virtually": the same as if
As far as the eye-brain system is
concerned, the effect is the same
as would occur if the mirror were
absent and the chess piece were located at the spot
labeled "virtual image".

48. Hall Mirror
● Useful to think in terms of images
“image” you
“real” you
mirror only
needs to be half as
high as you are tall. Your
image will be twice as far from you
as the mirror.

49. LEFT- RIGHT REVERSAL

50. Curved mirrors
● What if the mirror isn’t flat?
○ light still follows the same rules, with local surface normal
● Parabolic mirrors have exact focus
○ used in telescopes, backyard satellite dishes, etc.
○ also forms virtual image

51. Concave Mirrors
■ Curves inward
■ May be real or virtual
image

52. Focal Point:
The focal point of a curved mirror or
lens is the point at which incident rays
which are all parallel reflect off or refract
through the optical device converge. The
focal length (f) is the distance from this
convergence point to the mirror or lens.
MIRRORS AND LENSES

53. For a real object between f and the mirror, a
virtual image is formed behind the mirror.
The image is upright and larger than the
object.

54. For a real object between C and f, a real
image is formed outside of C. The image is
inverted and larger than the object.

55. For a real object at C, the real image is
formed at C. The image is inverted and the
same size as the object.

56. For a real object close to the mirror but outside
of the center of curvature, the real image is
formed between C and f. The image is inverted
and smaller than the object.

57. For a real object at f, no image is formed. The
reflected rays are parallel and never converge.
What size image is formed if
the real object is placed at
the focal point f?

58. Convex Mirrors
■ Curves outward
■ Reduces images
■ Virtual images
○ Use: Rear view mirrors,
store security…
CAUTION! Objects are closer than they
appear!

59. Refraction
● Light also goes through some things
○ glass, water, eyeball, air
● The presence of material slows light’s progress
○ interactions with electrical properties of atoms
● The “light slowing factor” is called the index of refraction
○ glass has n = 1.52, meaning that light travels about 1.5 times slower in glass
than in vacuum
○ water has n = 1.33
○ air has n = 1.00028
○ vacuum is n = 1.00000 (speed of light at full capacity)

60. n2 = 1.5
n1 = 1.0
A
B
Refraction at a
plane surface
● Light bends at interface between refractive indices
○ bends more the larger the difference in refractive index

61. Convex Lenses
Thicker in the center than edges.
○ Lens that
converges (brings
together) light
rays.
○ Forms real images
and virtual images
depending on
position of the
The Magnifier

62. Concave Lenses
● Lenses that are thicker at
the edges and thinner in the
center.
○ Diverges light rays
○ All images are
erect and reduced.
The De-Magnifier

63. How You See
● Near Sighted – Eyeball is too long,
and image focuses in front of the
retina
■ Near Sightedness – Concave lenses
expand focal length
● Far Sighted – Eyeball is too short,
so image is focused behind the
retina.
■ Far Sightedness – Convex lenses

64. Cameras, in brief
In a pinhole camera, the hole is so small that light hitting any particular point
on the film plane must have come from a particular direction outside the
camera
In a camera with a lens, the same applies that a point on the film plane
more-or-less corresponds to a direction outside the camera. Lenses have
the important advantage of collecting more light than the pinhole admits
pinhole
image at
film plane
object
image at
film plane
object
lens

65. STEPS FOR SOLVING MIRROR OR LENS EQUATION
PROBLEMS
Identify the given
quantities in the
problem including if the
optic device is a mirror
or lens, its focal point,
the image distance, the
object distance, the
magnification, the image
height, and/or the
object height
Determine which of the
equations
need to be used to solve
for the requested
information.
Solve the equations for
the desired variable(s).
Step 1 Step 2 Step 3
https://study.com/skill/learn/how-to-solve-mirror-lens-equation-problems-explanation.html

66. Mirror or Lens Equation:
The distance from the object to the
mirror or lens (do) and the distance from
the image formed by the optic device and
the device (di) are related to the focal
point by the equation:
MIRRORS AND LENSES

67. Sign Conventions: The positive or negative signs on many
of the numbers in optics equations are meaningful:
 A positive image distance indicates a real image
whereas a negative image distance indicates a virtual
image.
 A positive image height (or magnification) indicates an
upright image whereas a negative image height (or
magnification) indicates an inverted image.
 A positive focal length indicates a converging optic which
is either a concave mirror or a convex lens.
 A negative focal length indicates a diverging optic such
MIRRORS AND LENSES

68. Example Problem 1
A 5.0-cm tall candle is placed 10.0 cm away from a
convex mirror with a focal length of 15 cm. Determine
where the image forms and classify the image as either
real or virtual, inverted or upright, and larger, smaller, or
the same size.
Step 1: We will first determine all the known quantities. We
have been given the object height, the object distance, and
the focal length (which we know to be positive because the
mirror is convex):
Solving Mirror or Lens Equation Problems

69. Step 3: We will first solve the mirror equation for the image
distance. We can subtract both sides by , causing that
term to cancel from the left side:
We can then solve for image distance by taking the
reciprocal of both sides:
Solving Mirror or Lens Equation Problems

70. Plugging in known quantities, we get:
We can then plug this and the object distance into the magnification
equation to solve for the magnification of the system:
The image is formed 30 centimeters away from the mirror. The negative
sign on the image distance indicates the image is virtual. The positive sign
on the magnification indicates the image is upright. The fact that the
magnification is larger than 1 means that the image is larger than the
object (in this case, by a factor of 3). Thus, this is a virtual, upright,
larger image at a distance of 30 centimeters.
Solving Mirror or Lens Equation Problems

71. Example Problem 2
A 0.10-m tall light bulb is placed exactly 0.75 meters
from a diverging lens with a focal length of 0.25 m. Where
will the image form? What is the height of the image?
Step 1: We know the object distance, the focal length of
the lens, and the object height:
Solving Mirror or Lens Equation Problems

72. Step 2: In order to solve this problem for all the requested
information, we need both equations, although only need
the ratios of the distances and heights in the magnification
equation:
Solving Mirror or Lens Equation Problems

73. Step 3: We will first solve for the image distance again
using the lens equation solved for di as we did in the
previous problem:
Plugging in known quantities, we get:
Solving Mirror or Lens Equation Problems

74. We can use this to find the image height in the magnification
equation:
Multiplying both sides by the object height to cancel this from the
right side gives us:
The image is located 0.19 meters from the lens and has a height
of 0.025 meters.
Solving Mirror or Lens Equation Problems

75. AWESOME

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77. RESOURCES
Photos
● Cheerful friends with sparklers at night
● Crop hands with sparklers
Vector
● Bokeh lights wallpaper in gradient