1. Optics: Reflection, Refraction Mirrors and Lenses
05/25/2006
Optics: Reflection, Refraction
Optics
Mirrors and Lenses
Flat mirrors
Thin Lenses
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Lecture 16

2. Regular vs. Diffuse Reflection
Smooth, shiny surfaces have a regular reflection:
Rough, dull surfaces have a diffuse reflection.
Diffuse reflection is when light is scattered in different directions
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3. Optics: Reflection, Refraction
05/25/2006
Optics: Reflection, Refraction
Reflection
We describe the path of light as straight-line rays
Reflection off a flat surface follows a simple rule:
angle in (incidence) equals angle out (reflection)
angles measured from surface “normal” (perpendicular)
Laws of reflection
1 )The incident ray,
the reflected ray
and the normal
all lie in the same plane.
normal
incident ray
reflected ray
mirror  ˊ
2)The incident angel = the reflected angel
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Lecture 16

4. 24-1 Flat mirrors
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5. An object viewed using a flat mirror appears to be located behind the mirror, because to the observer the diverging rays from the source appear to come from behind the mirror
The image distance behind the mirror equals the object distance from the mirror
The image height h’ equals the object height h so that the lateral magnification
The image has an apparent left-right reversal
The image is virtual, not real!
1- Virtual images - light rays do not meet and the image is always upright or right-side-up“ and also it cannot be projected Image only seems to be there
Real images - always upside down and are formed when light rays actually meet
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6. example
If the angle of incidence of a ray of light is 42owhat is each of the following?
A-The angle of reflection (42o)
B-The angle the incident ray makes with the mirror (48o)
C-The angle between the incident ray and the reflected (90o)
T.Norah Ali Al-moneef king Saud university
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7. Now you look into a mirror and see the image of yourself.
a) In front of the mirror.
b) On the surface of the mirror.
C)Behind the mirror.
T.Norah Ali Al-moneef king Saud university
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8. Example A girl can just see her feet at the bottom edge of the mirror.
Her eyes are 10 cm below the top of her head.
(a) What is the distance between the girl and her image in the mirror?
 Distance = 150  2 = 300 cm
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9. Signs: Image size and magnification
images can be upright (positive image size h’) or inverted (negative image size h’)
Define magnification m = h’/h
Positive magnification: image orientation unchanged relative to object
Negative magnification: image inverted relative to object
l m l < 1 if image is smaller than object
l m l > 1 if image is bigger than object
l m l = 1 if image is same size as object
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10. 24-2 Thin Lenses
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11. A lens is a transparent material made of glass or plastic that refracts light rays and focuses (or appear to focus) them at a point
A converging lens will bend incoming light that is parallel to the principal axis toward the principal axis.Any lens that is thicker at its center than at its edges is a converging lens with positive f.
A diverging lens will bend incoming light that is parallel to the principal axis away from the principal axis.Any lens that is thicker at its edges than at its center is a diverging lens with negative f
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12. Rules For Converging Lenses
Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens.
Any incident ray traveling through the focal point on the way to the lens will refract through the lens and travel parallel to the principal axis.
An incident ray which passes through the center of the lens will in effect continue in the same direction that it had when it entered the lens.
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14. S- S s S- S S-
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15. The lens maker’s formula (lens in a medium)
( n lens = index of refraction of the lens material)
( nmedium= index of refraction of the medium)
The focal length f for a lens.
( n = index of refraction of the lens material)
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17. Ray Diagram for Converging Lens, S > f
The image is real
The image is inverted
The image is on the back side of the lens
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19. Ray Diagram for Converging Lens, S < f
The image is virtual
The image is upright
The image is larger than the object
The image is on the front side of the lens
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20. Object Outside 2F F 2F
Real; inverted; diminished
1. The image is inverted, i.e., opposite to the object orientation.
2. The image is real, i.e., formed by actual light on the opposite side of the lens.
3. The image is diminished in size, i.e., smaller than the object.
Image is located between F and 2F
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21. Object at 2F F 2F
Real; inverted; same size
1. The image is inverted, i.e., opposite to the object orientation.
2. The image is real, i.e., formed by actual light on the opposite side of lens.
3. The image is the same size as the object.
Image is located at 2F on other side
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22. Object Between 2F and F F 2F
Real; inverted; enlarged
1. The image is inverted, i.e., opposite to the object orientation.
2. The image is real; formed by actual light rays on opposite side
3. The image is enlarged in size, i.e., larger than the object.
Image is located beyond 2F
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23. Object at Focal Length F
Parallel rays; no image formed
When the object is located at the focal length, the rays of light are parallel. The lines never cross, and no image is formed.
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24. Object Inside F Virtual; erect; enlarged
1. The image is erect, i.e., same orientation as the object.
2. The image is virtual, i.e., formed where light does NOT go.
3. The image is enlarged in size, i.e., larger than the object.
Image is located on near side of lens
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25. Example. A magnifying glass consists of a converging lens of focal length 25 cm. A bug is 8 mm long and placed 15 cm from the lens. What are the nature, size, and location of image. F
S = 15 cm; f = 25 cm
S-= cm
The fact that S- is negative means that the image is virtual (on same side as object).
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26. . Example
Where must an object be placed to have unit magnification (
M = 1.00) (a) for a converging lens of focal length 12.0 cm ? (b) for a diverging lens of focal length 12.0 cm ? b a
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27. example
A person uses a converging lens that has a focal length of 12.5 cm to inspect a gem. The lens forms a virtual image 30.0 cm away. Determine the magnification. Is the image upright or inverted?
solution
Since
, the image is upright.
T.Norah Ali Al-moneef king Saud university
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28. example
A ray that starts from the top of an object and runs parallel to the axis of the lens, would then pass through the
a)principal focus of the lens
b)center of the lens
C)secondary focus of the lens
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29. Example 5: Derive an expression for calculating the magnification of a lens when the object distance and focal length are given.
From last equation: = -s M
Substituting for q in second equation gives . . .
Thus, . . .
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30. Diverging Thin Lens
Incoming parallel rays DIVERGE from a common point FOCAL
We still call this the point
Same f on both sides of lens
Negative focal length
Thinner in center
T.Norah Ali Al-moneef king Saud university
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31. Ray Diagrams for Thin Lenses – Diverging
For a diverging lens, the following three rays are drawn:
Ray 1 is drawn parallel to the principal axis and emerges directed away from the focal point on the front side of the lens
Ray 2 is drawn through the center of the lens and continues in a straight line
Ray 3 is drawn in the direction toward the focal point on the back side of the lens and emerges from the lens parallel to the principal axis
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32. Ray Diagram for Diverging Lens
The image is virtual
The image is upright
The image is smaller
The image is on the front side of the lens
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33. Sign Conventions for Thin Lenses
Quantity
Positive When
Negative When
Object locatio (s)
Object is in front of the lens
Object is in back of the lens
Image location (sˊ)
Image is in back of the lens
Image is in front of the lens
Image height (h’)
Image is upright
Image is inverted
R1 and R2
Center of curvature is in back of the lens
Center of curvature is in front of the lens
Focal length (f)
Converging lens
Diverging lens
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35. Example :
An object is placed 10 cm from a 15-cm-focal-length converging lens. Determine the image position and size (a) analytically, and (b) using a ray diagram. .
. An object placed within the focal point of a converging lens produces a virtual image.
Solution: a. The thin lens equation gives = -30 cm and M = 3.0. The image is virtual, enlarged, and upright.
b. See the figure. S’
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36. Example.
Where must a small insect be placed if a 25-cm-focal-length diverging lens is to form a virtual image 20 cm from the lens, on the same side as the object?
Since the lens is diverging, the focal length is negative. The lens equation gives = 100 cm. S
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39. The power of lens
The reciprocal of the focal length = the power of lens
If the focal length f is measured in meters then ;p measured in diopters if two lenses with focal length f1 and f2 placed next to each other are equivalent to a single lens with a focal length f satisfying
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40. Spherical Aberration
Results from the focal points of light rays far from the principle axis are different from the focal points of rays passing near the axis
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41. Spherical Aberration With SA SA free
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42. Chromatic Aberration
Different wavelengths of light refracted by a lens focus at different points
Violet rays are refracted more than red rays
The focal length for red light is greater than the focal length
for violet light
Chromatic aberration can be minimized by the use of a combination of converging and diverging lenses
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43. Multiple lenses can be used to improve aberrations
Spherical Aberration
Chromatic Aberration
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44. Lens Aberrations
Chromatic aberration can be improved by combining two or more lenses that tend to cancel each other’s aberrations. This only works perfectly for a single wavelength, however.
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45. An object is placed 6.0 cm in front of a convex thin lens of focal length 4.0 cm. Where is the image formed and what is its magnification and power?
s = 6.0 cm
f = 4.0 cm 1 s f = _ 1 s s’ f + = s’ 1 6 4 = - s’
= 12 cm s’
P = 1
0.04 m
= 25.0 D
M = - 12 / 6 = -2
Negative means real, inverted image
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46. 1 s s’ f + =
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48. Example 1. A glass meniscus lens (n = 1
Example 1. A glass meniscus lens (n = 1.5) has a concave surface of radius –40 cm and a convex surface whose radius is +20 cm. What is the focal length of the lens.
-40 cm
+20 cm
n = 1.5
R1 = 20 cm, R2 = -40 cm
f = 80.0 cm
Converging (+) lens.
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49. Example: What must be the radius of the curved surface in a plano-convex lens in order that the focal length be 25 cm?
R1=
R2=?
f = ?
R1 = , f= 25 cm
R2 = 0.5(25 cm)
R2 = 12.5 cm
Convex (+) surface.
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50. Example : What is the magnification of a diverging lens (f = -20 cm) the object is located 35 cm from the center of the lens? F
First we find q then M
s = cm
M =
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51. Example Real image, magnification = -1
An object is placed 20 cm in front of a converging lens of focal length 10 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image?
Real image,
magnification = -1
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52. Example
An object is placed 8 cm in front of a diverging lens of focal length 4 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image?
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53. 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.
Image is real,
inverted.
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54. 24(e). Given a lens with the properties (lengths in cm) R1 = +30, R2 = +30, s = +10, and n = 1.5, find the following: f, s and m. Is the image real or virtual? Upright or inverted? Draw 3 rays.
Real side
Virtual side R1 . F1 F2 p R2
Image is virtual, upright.
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55. Example
An object is placed 5 cm in front of a converging lens of focal length 10 cm. Where is the image? Is it upright or inverted? Real or virtual? What is the magnification of the image?
Virtual image, as viewed from the right, the light appears to be coming from the (virtual) image, and not the object.
Magnification = +2
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57. Summary: Lensmaker’s Equation+ -
R1 and R2 are interchangeable
R1, R2 = Radii
n= index of glass
f = focal length
R1 and R2 are positive for convex outward surface and negative for concave surface.
Focal length f is positive for converging and negative for diverging lenses.
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58. Summary of Math Approach
Lens Equation:
Magnification:
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59. Summary of Sign Convention
1. Object p and image q distances are positive for real and images negative for virtual images.
2. Image height y’ and magnification M are positive for erect negative for inverted images
3. The focal length f and the radius of curvature R is positive for converging mirrors and negative for diverging mirrors.
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60. Alternative Solutions
It might be useful to solve the lens equation algebraically for each of the parameters:
Be careful with substitution of signed numbers!