Lecture 14 Images Chapter 34. Law of Reflection Dispersion Snell’s Law Brewsters Angle Preliminary...
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Transcript of Lecture 14 Images Chapter 34. Law of Reflection Dispersion Snell’s Law Brewsters Angle Preliminary...
Lecture 14 Images Chapter 34
• Law of Reflection• Dispersion• Snell’s Law• Brewsters Angle
Preliminary topics before mirrors and lenses
•Law of Reflection•Dispersion•Snell’s Law•Brewsters Angle
Geometrical Optics:Study of reflection and refraction of light from surfaces
The ray approximation states that light travels in straight linesuntil it is reflected or refracted and then travels in straight lines again.The wavelength of light must be small compared to the size ofthe objects or else diffractive effects occur.
Law of Reflection
1A
B
θ I = θ R
θi
θr
Mirror
Drawing Normals
Fermat’s PrincipleUsing Fermat’s Principle you can prove theReflection law. It states that the path taken by light when traveling from one point to another is the path that takes the shortesttime compared to nearby paths.
Two light rays 1 and 2 taking different paths between points A and B and reflecting off a
vertical mirror
1
2
A
B Plane Mirror
Use calculus - method of minimization
t = 1C ( h1
2 + y2 + h22 + (w−y)2 )
dtdy
=2y
h12 + y2
+−2(w−y)
h22 + (w−y)2 )
=0
y
h12 + y2
=(w−y)
h22 + (w−y)2 )
sinθ I = sinθR
θ I = θR
Write down time as a function of yand set the derivative to 0.
Law of Refraction: Snells Law
€
n1 sinθ1 = n2 sinθ2
n1
n2
How do we prove it?
1
v1
sinθ1 =1v2sinθ2
Air 1.0Glass 1.33
t =(1v1
h12 + y2 +
1v2
h22 + (w−y)2 )
dtdy
=0
1v1sinθ I =
1v2sinθR
n1 sinθ I =n2 sinθR
JAVA APPLET
Show Fermat’s principle simulator
What allows you to see various colors when white light passes through a prism
Dispersion
How does a Rainbow work?
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 dose not change when n changes.
v = fλλ changes when medium changesf does not change when medium changes
Why is light totally reflected inside a fiber optics cable? Internal reflection
€
n1 sinθ1 = n2 sinθ2
€
(1.33)sinθ1 = (1.00)sin90 =1.00
When θ1 ≥sin−1 1
1.33 ≥48.75 deglight won't get out of the material
Fiber Cable
Same here
Corner Reflector?
Show Total Internal reflection simulator
Halliday, Resnick, Walker: Fundamentals of Physics, 7th Edition - Student Companion Site
What causes a Mirage
1.06
1.091.08
1.07 1.071.08
1.09
sky eye
Hot road causes gradient in the index of refraction that increasesas you increase the distance from the road
Index of refraction
Inverse Mirage Bend
47. In the figure, a 2.00-m-long vertical pole extends from the bottom of a swimming pool to a point 50.0 cm above the water. What is the length of the shadow of the pole on the level bottom of the pool?
Consider a ray that grazes the top of the pole, as shown in the diagram below. Here θ1 = 35o, l1 = 0.50 m, and l2 = 1.50 m.
θ2
θ1
l2
l1
L x
air
water
shadow
x is given by
x = l1tanθ1 = (0.50m)tan35o = 0.35 m.
The length of the shadow is x + L.
L is given by
L=l2tan θ2
Use Snells Law to find θ2
Snells Law Example
According to the law of refraction, n2sinθ2 = n1sinθ1. We take n1 = 1 and n2 = 1.33
oo
n55.25
33.1
35sinsin
sinsin 1
2
112 =⎟⎟
⎠
⎞⎜⎜⎝
⎛=⎟⎟
⎠
⎞⎜⎜⎝
⎛= −− θθ
L is given by
.72.055.25tan)50.1(tan 22 mmlL o === θ
The length of the shadow is L+x.
L+x = 0.35m + 0.72 m = 1.07 m.
θ2
θ1
l2
l1
L x
air
water
shadow
Calculation of L
Polarization by Reflection Brewsters Law
(1.)sinθB =nsinθr
θB +θr =90 we get 100% polarized reflected wavesin θB =nsin(90 −θB) =ncosθB
θB =tan−1n Brewsters Law
Plane Mirrors Where is the image formed
Mirrors and Lenses
Plane mirrors
Normal
Angle ofincidence
Angle of reflectioni = - p
Real sideVirtual side
Virtual image
eye
Object distance = - image distanceImage size = Object size
Problem: Two plane mirrors make an angle of 90o. How many images are there for an object placed between them?
object
eye
1
2
3
mirror
mirror
Problem: Two plan mirrors make an angle of 60o. Find all images for a point object on the bisector.
object
2
4
5,6
3
1
mirror
mirror
eye
Assuming no spinAssuming an elastic collisionNo cushion deformation
d
d
Using the Law of Reflection to make a bank shot
What happens if we bend the mirror?
i = - p magnification = 1
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
• 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
1
p+1i=1f
m=−ip
Concept Simulator/Illustrations
Halliday, Resnick, Walker: Fundamentals of Physics, 7th Edition - Student Companion Site
Spherical refracting surfaces
n1
p+n2
i=n2 −n1
r
Using Snell’s Law and assuming small Angles between the rays with the central axis, we get the following formula:
Apply this equation to Thin Lenses where the thickness is small compared to object distance, image distance, and
radius of curvature. Neglect thickness.
Converging lens
Diverging lens
Thin Lens Equation
€
1
f=
1
p+
1
i
Lensmaker Equation
€
1
f= (n −1)(
1
r1−
1
r2
)
What is the sign convention?
Lateral Magnification for a Lensm =−
ip
Sign Convention
p
Virtual side - V Real side - R
i
Light
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.
r2r1
Rules for drawing rays to locate images from a lens
•A ray initially parallel to the central axis will pass through the focal point.
•A ray that initially passes through the focal point will emerge from the lens parallel to the central axis.
• A ray that is directed towards the center of the lens will go straight through the lens undeflected.
Example of drawing images
24(b). 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 3 rays.
pfi
111−=
€
m =′ y
y= −
i
p
€
1
i=
1
5−
1
10= +
1
10
Image is real, inverted.
. .F1 F2p
Virtual side Real side
€
m = −10
10= −1
€
i = +10 cm
Example
24(e). 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.
( ) ⎟⎟⎠
⎞⎜⎜⎝
⎛−−=
21
111
1
rrn
f
( )30
1
30
1
30
115.1
1=⎟
⎠
⎞⎜⎝
⎛−
−−=f
cmf 30=
pfi
111−=
15
1
10
1
30
11−=−=
i
cmi 15−=
€
m =′ y
y= −
i
p
€
m = −−15
10= +1.5
Image is virtual, upright.
Virtual side Real side
r1. .F1 F2
pr2
27. 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 lensSeparately.
f1
f1
Lens 1 Lens 2
f2
f2
10
40
+20 -15
f1
f1
Lens 1 Lens 2
f2
f2
10
40
+20 -15
Ignoring the diverging lens (lens 2), the image formed by theconverging lens (lens 1) is located at a distance
€
1
i1=
1
f1
−1
p1
=1
20cm−
1
40cm. i1 = 40cm
40
This image now serves as a virtual object for lens 2, with p2 = - (40 cm - 10 cm) = - 30 cm.
30
Since m = -i1/p1= - 40/40= - 1 , the image is inverted
€
1
i2
=1
f2
−1
p2
=1
−15cm−
1
−30cm i2 = −30cm.
Thus, the image formed by lens 2 is located 30 cm to the left of lens 2. It is virtual (since i2 < 0).
f1
f1
Lens 1 Lens 2
f2
f2
10
40
+20 -15
40
30
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
Magnifying lensCompound microscopeRefracting telescope