Physics 212 Lecture 26: Lenses
Physics 212 Lecture 26, Slide 1
Music Who is the Artist? A) B) C) D) E)
Ramblin’ Jack Elliott Arlo Guthrie Pete Seeger Phil Ochs U. Utah Phillips
Why? Last time I had to talk about “circles & arrows” 46th anniversary of Arlo being arrested for litterin after Thanksgiving See the movie !! Physics 212 Lecture 26, Slide 2
Your Comments “This stuff is so hard!!!!” “Thank you for easier concepts at the very end of the semester.”
YOU CAN !! Just look in the mirror
“I just don't know what is virtual image? We can't see it right?” “I felt like there were a lot of different versions of the magnification and lense equations shown on the prelecture. Which one should we use?”
“I want to discuss light diffraction and what happens when 2 or 3 converging lenses are placed in front of one another.?” “how does this apply to glasses?” “Did you know I was able to bend my self into a convergent lens and make myself completely clear so I could refract light to a focal point? I made a spectacle of myself.”
Always differing opinions??
There are only two equations! They work for both lenses and mirrors You do have to know the CONVENTIONS We will discuss systems of lenses and the human eye in Lecture 28
“If you put this comment on the power point I will give you a box of girl scout cookies... You know you want some thin mints.” “TURKEY TURKEY TURKEY is all i can think of. HAPPY Thanksgiving EVERYone!!!!!.”
Remember: Deadline for signing up for conflict is Tomorrow 10pm!
Physics 212 Lecture 26, Slide 3
Refraction Snell’s Law n1sin(θ1) = n2sin(θ2) θ1
n1 n2 θ2
That’s all of the physics – everything else is just geometry! Physics 212 Lecture 26, Slide 4
air
water
θi
θi glass
θ2
1.3
θ2
1.5
glass 1.5
Case I
Case II
In Case I light in air heads toward a piece of glass with incident angle θi In Case II, light in water heads toward a piece of glass at the same angle. In which case is the light bent most as it enters the glass?
I or II or Same (A)
(B)
(C)
The angle of refraction in BIGGER for the water – glass interface:
n1sin(θ1) = n2sin(θ2)
sin(θ2)/sin(θ1) = n1/n2
Therefore the BEND ANGLE (θ1 – θ2) is BIGGER for air – glass interface Physics 212 Lecture 26, Slide 5
Checkpoint 2 What happens to the focal length of a converging lens when it is placed under water? A. increases B. decreases C. stays the same
The rays are bent more from air to glass than from water to glass Therefore, the focal length in air is less than the focal length in water
We can see this also from Lensmaker’s Formula
50 40 30 20 10 0
Physics 212 Lecture 26, Slide 6
Object Location • Light rays from sun bounce off object and go in all directions – Some hits your eyes
We know object’s location by where rays come from.
We will discuss eyes in lecture 28…
Physics 212 Lecture 26, Slide 7
Waves from object are focused by lens
Physics 212 Lecture 26, Slide 8
Two Different Types of Lenses
Physics 212 Lecture 26, Slide 9
Converging Lens: Consider the case where the shape of the lens is such that light rays parallel to the axis of the mirror are all “focused” to a common spot a distance f behind the lens:
f
f Physics 212 Lecture 26, Slide 10
Recipe for finding image: 1) Draw ray parallel to axis refracted ray goes through focus 2) Draw ray through center refracted ray is symmetric
object
f image
You now know the position of the same point on the image Physics 212 Lecture 26, Slide 11
S > 2f
image is: real inverted smaller
Example 1 1 1 + = S S′ f S′ M =− S
object
f image
f>0 S>0
S’ > 0 Physics 212 Lecture 26, Slide 12
S=f
Example 1 1 1 + = S S′ f
image is: at infinity…
S′ M =− S
object
f
f>0
S>0 Physics 212 Lecture 26, Slide 13
Example 0
1 1 1 + = S S′ f
image is: virtual upright bigger
S′ M =− S
image
f object
S>0 S’ < 0
f>0 Physics 212 Lecture 26, Slide 14
Diverging Lens: Consider the case where the shape of the lens is such that light rays parallel to the axis of the lens all diverge but appear to come from a common spot a distance f in front of the lens:
f Physics 212 Lecture 26, Slide 15
Example image is: virtual upright smaller
1 1 1 + = S S′ f S′ M =− S
object
f
image
f<0 S>0
S’<0
Physics 212 Lecture 26, Slide 16
Executive Summary - Lenses: S > 2f
real inverted smaller
2f > S > f
real inverted bigger
f >S>0
virtual upright bigger
S >0
virtual upright smaller
converging
1 1 1 + = S S′ f
diverging
f
S′ M =− S
f
Physics 212 Lecture 26, Slide 17
It’s always the same: 1 1 1 + = S S′ f
S′ M =− S
You just have to keep the signs straight:
The sign conventions
S: positive if object is “upstream” of lens S’ : positive if image is “downstream” of lens f: positive if converging lens
Physics 212 Lecture 26, Slide 18
Checkpoint 1a
A B C D
60
Image on screen
s′ > 0
MUST BE REAL
50 40 30 20
s′ M =− <0 s
MUST BE INVERTED
10 0
Physics 212 Lecture 26, Slide 19
Checkpoint 1b
A converging lens is used to project the image of an arrow onto a screen as shown above. A piece of black tape is now placed over the upper half of the lens. Which of the following is true? A. Only the lower half of the object (i.e. the arrow tail) will show on the screen B. Only the upper half of the object (i. e. the arrow head) will show on the screen C. The whole object will show on the screen
“The rays transmitting the head of the arrow will be blocked so only the tail of the arrow will be shown. “
50 40
“the image inverts itself, and thus blocking the top half of the lens would block the bottom half of the source”
30 20
“The brightness of the image will be halved because half of 10 the rays from each point on the object to each point on the 0 image are blocked, but the whole image will still be there.“Physics 212
Lecture 26, Slide 20
object
image
object
image
Cover top half of lens Light from top of object
Cover top half of lens Light from bottom of object
What’s the Point? The rays from the bottom half still focus The image is there, but it will be dimmer !! A converging lens is used to project the image of an arrow onto a screen as shown above. A piece of black tape is now placed over the upper half of the lens. Which of the following is true? A. Only the lower half of the object (i.e. the arrow tail) will show on the screen B. Only the upper half of the object (i. e. the arrow head) will show on the screen Physics 212 Lecture 26, Slide 21 C. The whole object will show on the screen
Calculation A magnifying glass is used to read the fine print on a document. The focal length of the lens is 10mm. At what distance from the lens must the document be placed in order to obtain an image magnified by a factor of 5 that is NOT inverted?
• Conceptual Analysis • Lens Equation: 1/s + 1/s’ = 1/f • Magnification: M = -s’/s • Strategic Analysis • Consider nature of image (real or virtual?) to determine relation between object position and focal point • Use magnification to determine object position
Physics 212 Lecture 26, Slide 22
A magnifying glass is used to read the fine print on a document. The focal length of the lens is 10mm. At what distance from the lens must the document be placed in order to obtain an image magnified by a factor of 5 that is NOT inverted?
Is the image real or virtual? (A) REAL
(B) VIRTUAL A virtual image will be upright
A real image would be inverted
h
h’ f
h’
h f
Physics 212 Lecture 26, Slide 23
A magnifying glass is used to read the fine print on a document. The focal length of the lens is 10mm. At what distance from the lens must the document be placed in order to obtain an image magnified by a factor of 5 that is NOT inverted?
How does the object distance compare to the focal length? (A)
s< f
Lens equation
(B)
s = f
s > f
(C)
1 1 1 = − s′ f s
s′ =
fs s− f
Virtual Image fl s’ < 0 Real object fl s > 0 Converging lens fl f > 0
h’
s’
h
s
f
s− f <0 Physics 212 Lecture 26, Slide 24
A magnifying glass is used to read the fine print on a document. The focal length of the lens is 10mm. At what distance from the lens must the document be placed in order to obtain an image magnified by a factor of 5 that is NOT inverted?
s′ =
fs s− f
What is the magnification M in terms of s and f? (A) M =
s− f f
(B) M =
Lens equation:
Magnification equation:
1 1 1 = − ′ s f s
s′ M =− s
fs s′ = s− f
f −s f
(C) M =
h’
M=
−f s− f
s’
−f s− f
(D) M =
f s− f
h
s
f
Physics 212 Lecture 26, Slide 25
A magnifying glass is used to read the fine print on a document. The focal length of the lens is 10mm. At what distance from the lens must the document be placed in order to obtain an image magnified by a factor of 5 that is NOT inverted?
(A) 1.7mm
(B) 6mm
(C) 8mm
−f M= s− f
(D) 40 mm
(E) 60 mm
M = +5 f = +10 mm M=
−f s− f
s=
M − 1) ( s= f M
h’
4 f = 8 mm 5
h
s′ = − sM = −40 mm
f Physics 212 Lecture 26, Slide 26
Follow Up Suppose we replace the converging lens with a diverging lens with focal length of 10mm. If we still want to get an image magnified by a factor of 5 that is NOT inverted, how does the object sdiv compare to the original object distance sconv?
(A) sdiv < sconv
(B) sdiv = sconv
(C) sdiv > sconv
EQUATIONS M=
−f s− f
(D) sdiv doesn’t exist
PICTURES s= f
M −1 M
h h’
M = +5 f = −10 mm
4 s = f = −8 mm 5
s negative fl not real object
s
f
s’
Draw the rays: s’ will always be smaller than s Magnification will always be less than 1 Physics 212 Lecture 26, Slide 27
Follow Up Suppose we replace the converging lens with a diverging lens with focal length of 10mm. M=
What is the magnification if we place the object at s = 8mm?
(A) M =
1 2
(B) M = 5
EQUATIONS M=
−f s− f
s = 8 mm f = −10 mm
(C)
M =
3 8
(D)
M=
5 9
(E)
−f s− f
M=
4 5
PICTURES
−10 10 5 M =− = = 8 − (−10) 18 9
h h’ f
Physics 212 Lecture 26, Slide 28