pinhole camera investigation

Post on 24-Feb-2016

68 Views

Category:

Documents

1 Downloads

Preview:

Click to see full reader

DESCRIPTION

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. - PowerPoint PPT Presentation

TRANSCRIPT

Pinhole Camera Investigation

By Oscar G.M and David Mayer

IntroductionPurposes: • 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 AnalysisImage Height vs. Object Distance

1. 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/m6. 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 = HoSo

Slope= Ho* di

Error CalculationsAccepted 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 AnalysisImage Height vs. Object Height

1. Image Height Hi Object Height Ho

2. Hi α Ho

3. Hi = k* Ho

4. k = ∆ Hi / ∆ Ho

5. k = 0.812 mm/mm6. Hi = 0.812 mm/mm * Ho

Let’s look at those units once more… mm/mm 1This looks like a ratio…our constants were object distance: 0.20 m, image

distance: 0.16 mConstant / 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 AnalysisImage Height vs. Image Distance

1. Image Height Hi Image Distance di

2. Hi = α di

3. Hi = k*di

4. k = ∆ Hi / ∆ di

5. k = 0.100 m/m6. 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 distanceSo

Slope= Ho/ do

Error CalculationsAccepted Value=0.0230m/0.200m=0.115 Experimental Value=0.100

Pinhole

Light Source with rays

ObjectHo

do

Hi

di

The Basic Principle

Ho

Hi

di

Original

do

Part 1: Changing Object Distance

Ho

do

Hi

di

Original

Part 2: Changing Object Height

Ho

do diHi

OriginalPart 3: Changing Image Distance

2 More Ways to See This RelationshipJoint Variation:

Hi α 1/ do

Hi α Ho

Hi α di

Hi α Ho * di /do

Hi = Ho * di /doThere was no constant of proportionality

Geometry

doHi

di

Ho

θ

θ

More Pinholes

Bigger Diameter Pinhole

Bigger Hole = Brighter, Blurrier Image

The End

top related