Pinholes-David and Oscar

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Pinhole Camera Investigation
By Oscar G.M and David Mayer
Introduction
Purposes:
• To investigate the relationship between object distance
and image height while maintaining a constant image
distance, and a constant object height.
• To investigate the relationship between object height
and image height while maintaining a a constant
object distance, and a constant image distance.
• To investigate the relationship between image distance
and image height while maintaining a constant object
height, and a constant object distance.
• To observe the effects of increasing the diameter of the
pinhole, and creating more than one pinhole.
Hypotheses
• If we decrease the object distance, then the
image height will increase.
• If we increase the object height, then the
image height will increase as well.
• If we increase the image distance then the
object distance will increase as well.
Equipment
Setup
Part 1: Image Height vs. Object Distance
Image Distance: 0.16 m
Object Height: 0.023m
Mathematical Analysis
1.
Image Height vs. Object Distance
Image Height  Hi Object Distance  do
2. Hi α 1/ do
3. Hi = k* 1/ do
4. k = ∆ Hi / ∆ 1/ do
5. k = 0.00399 m/1/m
6. Hi = 0.00399 m/1/m * 1/ do
Let’s look at those units… m/(1/m)  m*(m/1)  m2
Hmm, m2 … what were our constants again? Image distance: 0.16 m, object
Height: 0.023m. m*m  m2 .
0.023m * 0.16 m = 0.00368 m2 .
Meaning?
0.023m * 0.16 m = image distance * object height
Image distance = di
And
Object height = Ho
So
Slope= Ho* di
Error Calculations
Accepted Value=0.16m*0.023m=0.00368m2
Experimental Value=0.00400m2
Part 2: Image Height vs. Object Height
Object Distance: 0.20 m
Image Distance: 0.16 m
Mathematical Analysis
1.
Image Height vs. Object Height
Image Height  Hi Object Height  Ho
2. Hi α Ho
3. Hi = k* Ho
4. k = ∆ Hi / ∆ Ho
5. k = 0.812 mm/mm
6. Hi = 0.812 mm/mm * Ho
Let’s look at those units once more… mm/mm  1
This looks like a ratio…our constants were object distance: 0.20 m, image
distance: 0.16 m
Constant / Constant = 0.812
Meaning?
The only way for this to occur is if we divide the smaller
number by the larger number to give us something
less than 1.
0.16 m / 0.200 m = 0.800
Slope= di/do
Error Calculations
Accepted Value=0.16m/0.2m=0.800
Experimental Value=0.812
Part 3: Image Height vs. Image Distance
Object Height: 0.023 m
Object Distance: 0.200 m
Mathematical Analysis
1.
Image Height vs. Image Distance
Image Height  Hi Image Distance  di
2. Hi = α di
3. Hi = k*di
4. k = ∆ Hi / ∆ di
5. k = 0.100 m/m
6. Hi = 0.100 m/m *di
Looks like a ratio once more…our constants were object height: 0.023 m,
object distance: 0.200 m
Constant / Constant = 0.100
Meaning (last cheesy time)?
0.023m / 0.200 m = object height / object distance
So
Slope= Ho/ do
Error Calculations
Accepted Value=0.0230m/0.200m=0.115
Experimental Value=0.100
The Basic Principle
Pinhole
Ho
Object
Hi
do
Light Source with rays
di
Part 1: Changing Object
Distance
Original
Ho
Hi
do
di
Part 2: Changing Object
Height
Original
Ho
Hi
do
di
Part 3: Changing Image
Distance
Original
Ho
do
di
Hi
2 More Ways to See This
Relationship
Joint Variation:
Hi α 1/ do
Hi α Ho
Hi α di
Hi α Ho * di /do
Hi = Ho * di /do
There was no constant of proportionality
Geometry
Ho
θ
θ
Hi
do
di
More Pinholes
Bigger Diameter Pinhole
Bigger Hole = Brighter, Blurrier Image
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