Pixels, resoultion and light sensitive chips.

[greener1993]

I have the answer to all theses questions beacuse they are from a book but don't quite understand how to get it. Nice and good to cheat on my homework but not good in exam :S i am just so confused. i understand the resoultion is the smallest thing you can see in the picture which in most cause is a pixel which also in most cases is 1MM long. However i can't seem to gather how to work out questions. Why not use the book? beacuse i swear to god it is useless. OCR advancing physics book. Its more of a history book about physics than anything that will get us a good grade.

LS= light sensitive.
Red = Word I do not understand or there relationship to the question

1)a digital camera has a lens of focus length 50mm and produces an image of 1280x960 pixels. A human face (270x200)that is 1.5mfrom the lens fills the picture Of the light sensitive chip.
A) Estimate the dimension of the picture area of the LS chip in the camera.
B) Estimate the dimensions of one picture element in the LS chip.
C) estimate the scale of features on the face that can be resolved in the picture. could a person eyelashes be resolved?



2) Estimate how many pixels are needed in a satellite image of a town, if parked cars are to be resolved in the image.
Information : town = 10km2
lenth of car = 3m
resoulution = 3m

3)Estimate how many pixels are needed in an ultrasound scan image of a full-term foetus, if the baby’s fingers are to be distinguished
Information : Width of finger is 3mm
length of foetus a few hundred mm

4) Estimate the dimensions of an area of space imaged around a star, if an Earth-sized planet is just detectable in a 100 M pixel image.
Informtaion : Earth is about 10 000 km in diameter.

I just need it explaining well. Would be nice if you could include any equations and step by step instructions

Thank you :)

 

[jbsteele]

A) Estimate the dimension of the picture area of the LS chip in the camera. The dimension of the picture area of the LS chip in the camera is 1280 x 960 pixels, which is roughly equivalent to 1.2 million square pixels. B) Estimate the dimensions of one picture element in the LS chip. The dimensions of one picture element in the LS chip is calculated by dividing the total number of pixels in the LS chip by the area of the picture, which is 1280 x 960. This gives us a result of 1.2 million/1280 x 960, which is roughly equivalent to 0.00140625 mm (1.4 microns). C) estimate the scale of features on the face that can be resolved in the picture. could a person eyelashes be resolved? The scale of features on the face that can be resolved in the picture is determined by the size of the picture element in the LS chip. In this case, each picture element is 0.00140625 mm (1.4 microns) wide. Therefore, a person's eyelashes would not be able to be resolved in the picture, as they are usually much larger than this size. 2) Estimate how many pixels are needed in a satellite image of a town, if parked cars are to be resolved in the image.Information : town = 10km2 lenth of car = 3m resoulution = 3mThe number of pixels needed in a satellite image of a town can be calculated by dividing the area of the town (10 km2) by the resolution of the image (3 m). This gives us a result of 3333 x 3333 pixels or 11.1 million pixels. 3) Estimate how many pixels are needed in an ultrasound scan image of a full-term foetus, if the baby’s fingers are to be distinguishedInformation : Width of finger is 3mm length of foetus a few hundred mmThe number of pixels needed in an ultrasound scan image of a full-term foetus can be calculated by dividing the length of the foetus (a few hundred mm) by the width of the finger (3 mm). This gives us a result of 100 x 100

 

[kerimek]

I understand your struggle with understanding these concepts and questions. Let me break down each question and explain it in a way that will hopefully make it clearer for you.

1) In this question, we are given information about a digital camera with a lens of 50mm focus length and a resolution of 1280x960 pixels. We are also given the size of a human face (270x200) and its distance from the lens (1.5m). Based on this information, we need to estimate the dimension of the light sensitive chip in the camera, the dimensions of one picture element (pixel) in the chip, and the scale of features that can be resolved in the picture.

A) To estimate the dimension of the light sensitive chip, we need to use the formula: Chip Dimension = (Picture Dimension x Distance to Object) / Focal Length. In this case, the picture dimension is 1280x960 pixels and the distance to the object (human face) is 1.5m. So, the chip dimension would be (1280x960 x 1.5) / 50 = 38.4mm x 28.8mm.

B) To estimate the dimensions of one pixel in the chip, we can divide the chip dimension by the number of pixels in the picture. So, one pixel would be 38.4mm / 1280 = 0.03mm long.

C) To estimate the scale of features that can be resolved in the picture, we need to use the formula: Resolving power = 1.22 x (Wavelength of light / Aperture). In this case, the aperture would be the size of one pixel, which we calculated to be 0.03mm. The wavelength of light would be around 500nm (green light). So, the resolving power would be 1.22 x (500nm / 0.03mm) = 20.33 micrometers. This means that features on the face that are larger than 20.33 micrometers can be resolved in the picture. So, a person's eyelashes, which are typically around 10-12mm, would be able to be resolved in the picture.

2) In this question, we need to estimate the number of pixels needed in a satellite image of a town in order to resolve parked cars. We are given the size of the town, the length of