Upload
eleanore-owens
View
219
Download
0
Tags:
Embed Size (px)
Citation preview
Lecture 14 Images Chp. 35
• Opening Demo• Topics
– Plane mirror, Two parallel mirrors, Two plane mirrors at right angles
– Spherical mirror/Plane mirror comparison in forming image– Spherical refracting surfaces– Thin lenses– Optical Instruments
Magnifying glass, Microscope, Refracting telescope• Warm-up problem
Geometrical Optics:Study of reflection and refraction of light from surfaces using the ray approximation.
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.
Reflection Refraction: Snells Law
€
n1 sinθ1 = n2 sinθ2
€
θ1 = θ1'
n1
n2
Using Fermat’s Principle you can prove theabove two laws. 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.
Plane mirrors
Normal
Angle ofincidence
Angle of reflection
i = - p
Real side Virtual side
Virtual image
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
How tall must be a mirror be for a person to see his entire reflection in it. Let H be the height of a man and let l be the height of the mirror?
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.
Thin Lenses: thickness is small compared to object distance, image distance, and radius of curvature.
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?
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 o the V side.
r2r1
Rules for drawing rays to locate images
•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.
Show Fig 35-14 figure overhead
Then go to example
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
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