# pinhole camera investigation

Post on 24-Feb-2016

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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. - PowerPoint PPT PresentationTRANSCRIPT

Pinhole Camera Investigation

Pinhole Camera InvestigationBy Oscar G.M and David MayerIntroductionPurposes: 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. HypothesesIf 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 DistanceImage Height Hi Object Distance doHi 1/ doHi = k* 1/ dok = Hi / 1/ dok = 0.00399 m/1/mHi = 0.00399 m/1/m * 1/ do

Lets 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 = HoSoSlope= 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 mImage Distance: 0.16 m

Mathematical AnalysisImage Height vs. Object Height Image Height Hi Object Height HoHi HoHi = k* Hok = Hi / Hok = 0.812 mm/mmHi = 0.812 mm/mm * Ho

Lets look at those units once more mm/mm 1This looks like a ratioour 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 = 0.800Slope= 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 mObject Distance: 0.200 m

Mathematical AnalysisImage Height vs. Image DistanceImage Height Hi Image Distance diHi = diHi = k*dik = Hi / dik = 0.100 m/mHi = 0.100 m/m *di

Looks like a ratio once moreour constants were object height: 0.023 m, object distance: 0.200 mConstant / Constant = 0.100

Meaning (last cheesy time)?0.023m / 0.200 m = object height / object distanceSoSlope= Ho/ do

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

PinholeLight Source with raysObjectHodoHidiThe Basic PrincipleHoHidiOriginaldoPart 1: Changing Object DistanceHodoHidiOriginalPart 2: Changing Object Height HododiHiOriginalPart 3: Changing Image Distance 2 More Ways to See This RelationshipJoint Variation:Hi 1/ doHi HoHi diHi Ho * di /doHi = Ho * di /doThere was no constant of proportionality

GeometrydoHidiHo

More Pinholes

Bigger Diameter Pinhole

Bigger Hole = Brighter, Blurrier ImageThe End

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