Calculating Image Height from Magnification: How to Solve

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To determine the height of an image formed by a converging lens, one can use the magnification formula M = hi/ho, where hi is the height of the image and ho is the height of the object. A ray diagram can visually illustrate the relationship, but it is not the only method; trigonometry can also be applied. By using the object distance and height to form a right triangle, and the image distance and height for a similar triangle, one can calculate the image height mathematically. If two variables in the magnification equation are known, the image height can be derived without drawing a diagram. Thus, while ray diagrams are helpful, they are not strictly necessary for calculating image height.
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my question is how do you determine the height of the image do you have to draw the ray diagram first to scale and then measure physically with a ruler or is there a way to mathematically do it



the formula for calculating magnification of an object is
M=hi/ho (magnification = height of image/height of object)



so here is a sample problem
a 1.5cm high object is 8.0 cm from a converging lens of focal length 2.5 cm
for this the object height is given its 1.5cm and the image height is -0.70cm how do you get that by drawing?
 
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You should try doing the ray diagram if you don't see why the image height is negative.

This should hint you that there are two cases when the image height is inverted when dealing with converging lens, and it depends on the distance of the focal length.
 
There will be similar triangles using geometrical optics. You can use the thin lens equation to find the image distance and the object height and distance to find the angle. Then using trigonometry one can find the image height.
 
thanks...but

that was not what i was asking i know why the image is inverted what i am asking is that the only way to determine the image height is through the ray diagram and i know the um thin lens equation but same problem so all in all i have to do the ray diagram no matter what?
 
You can use trigonometry. The object distance and object height are two sides of a right angled triangle. The image distance and image height will be two sides of a similar triangle. You can compute the image height knowing these facts through simple trigonometry.
 
Well obviously if you have any two of the variables in the magnification equation, that's another way of finding one of the heights. In that case you don't need to draw the diagram.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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