Problem 25.07 Part A A certain digital camera having a lens with tocal length 7.
ID: 1793109 • Letter: P
Question
Problem 25.07 Part A A certain digital camera having a lens with tocal length 7.50 cm focuses on an object 1.65 tall that is 4.70 m from the lens How far must the lens be from the sensor aray? Express your answer in centimeters to three significant figures. cm Submit My Answers Give Up Part B How tall is the image on the sensor array? Express your answer in centimeters to three significant figures. Submit My Answers Give Up Part C Is the image on the photocells upright or inverted? Real or virtual? t and real. The image is inverted and real. The image is upright and virtual The image is inverted and virtual. Submit My Answers Give Up Part D A SLR digital camera often has pixeIs measuring 8.00 m 8.00 m How many such pixels does the height of this imesge cover? Express your answer in number of pixels to two significant figures. pixels Submit My Answers Give UpExplanation / Answer
To work this problem, best to convert all measurements to the same unit.
I choose millimeters.
You can construct a triangle; distance to subject is the height of the triangle. The subject’s height is the base of the triangle.
Height = 1,650mm
Distance = 4,700mm
The ratio of subject’s height to subject distance is 1650 ÷ 4700 = 0.35106
The camera lens projects and image onto film or digital imaging chip (onto the focal plane).
You can construct a triangle using the focal length of the lens as the height value. Thus the triangle height is 7.5cm X 10 =75mm
The lens projects an image of subject at the focal plane. = 7.50cm (a)
The height of this image is unknown however we can construct a congruent triangle.
The height of the height of the triangle is 75mm
The base of the triangle is the image height which is unknown.
However we can calculate the image height using the ratio value above.
The image height is 75 X 0.35106 = 26.3298mm = 2.63 cm (b)
Now we know the image height at the focal plane.
The image is upright. (c)
How many pixels needed to span this length?
Now the pixels measure 8 micrometers = 8um = 0.008mm
How many pixels needed to cover the span?
Answer = 26.3298 ÷ 0.008 = 3300 pixels. (d)
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