A Few Question On Lens and Magnification

[shimizua]

Homework Statement

So i thought it was Real, magnified, and inverted but just wanted to check. Then also i wanted to know this
here is the picture http://www.gatewaytosedona.com/image/articles4/2121/EST_Jupiter_416w.jpg

Look up the size of Jupiter and the distance to Jupiter
when it is closest to Earth. Use the lens formula to
determine the magnification of the image. How many
pixels in diameter is Jupiter’s image on the CCD? How
large a square on the surface of Jupiter does one pixel in
the image represent?

so the size of Jupiter is 1.9x10^27 kg
distance is 591 million km
and the diameter of Jupiter is 143,000,000 m
and the diameter of the image is .034 m

so to find the magnification i did m=hi/ho and go 2.38x10^-10, but i just didnt think it shouldbe so small since i thought it was magnified. but if i am wrong with the first part then that could be right
So then for the next with how man pixels, it would just be 1024 pixels right?
Then the last part would just be 143,000,000m/1024pixels and got 139648.44

Homework Equations

already kinda put the equations i used

The Attempt at a Solution

just really wanted to make sure i was doing it all right

 

[anathema]

Thank you for reaching out for clarification on your assignment. Based on the information and equations you provided, your calculations seem to be correct. Jupiter's image on the CCD would be approximately 1024 pixels in diameter, and one pixel in the image would represent a square on the surface of Jupiter of approximately 139,648.44 meters.

As for the magnification, it is important to note that the magnification of the image is small because the distance between Earth and Jupiter is large. Magnification is calculated by dividing the image distance (hi) by the object distance (ho), so the larger the object distance, the smaller the magnification. In this case, the distance between Earth and Jupiter is 591 million kilometers, which would result in a small magnification.

I hope this helps clarify any confusion and that you are able to complete your assignment successfully. Keep up the good work in your studies of science!

 

[nlightner]

I would like to provide a response to your questions about lenses and magnification. Firstly, your understanding that images formed by lenses are real, magnified, and inverted is correct. As for your calculations, I would like to point out a few things.

The size of Jupiter is not measured in kilograms, but rather in meters. Its diameter is approximately 143,000 kilometers, not meters. This is an important distinction to make when using the lens formula, which is based on distances in meters.

Next, the distance to Jupiter varies as it orbits the sun, so you should specify at what point in its orbit you are referring to. For example, when it is closest to Earth, it is approximately 591 million kilometers away.

Using these corrected values, the magnification of the image can be calculated as follows:
m = -di/do = -dobject/dimage = (591 million km)/(143,000 km) = 4130

This means that the image of Jupiter is magnified 4130 times compared to its actual size. This is a significant amount of magnification, and it is not uncommon for astronomical images to have such high magnification ratios.

As for the number of pixels in the image, it would depend on the resolution of the camera used to capture the image. Assuming a standard resolution of 1024 x 1024 pixels, then yes, there would be 1024 pixels in diameter for the image of Jupiter.

Finally, to calculate the size of one pixel on the surface of Jupiter, you would divide the diameter of Jupiter (143,000 km) by the number of pixels (1024). This gives a value of approximately 139.6 km per pixel.

I hope this helps clarify your understanding of lenses and magnification, and how they apply to the image of Jupiter provided. Keep up the good work in your studies!