
#1
Dec1512, 04:21 AM

P: 106

We're learning about magnification and they say how magnification is the ratio of the visual angle while looking through the instrument to that with the naked eye and then they say,
For small angles, magnification is defined as, m= height of the image/height of the object Why only for small angles, I thought this works for any angle? Is it an approximation. I thought that height of image/height of object was the definition of magnification in the first place, how can it not apply for large angle or is this only when we're looking for magnification as it appears to our eye? 



#2
Dec1512, 04:37 AM

P: 106

And also, in a simple microscope where they use a single convex mirror and an object within the focus of the convex lens so that it produces a virtual image, there they use the formula v/u to get the magnification, but I feel that even though the image is bigger, it's also behind which reduces the visual angle, so magnification (in terms of how much bigger the object APPEARS to us) won't just be v/u.




#3
Dec1512, 04:51 AM

P: 106

http://www.citycollegiate.com/magnifying_glass.htm This site contains more detailed info on what they were talking about. But I don't get the math. Just because alpha and beta are small angles why should alpha be equal to tan alpha?




#4
Dec1512, 11:32 AM

P: 197

magnification of an image
Hmm.. It's true. You can check it with your calculator. Sine of a small angle is equal to that angle (in radians). The best way to see that aproximation is to look at trigonometric circle. You reed the value of sine of an arc on Y axis. You can see how as angle aproches 0, the length of an arc (the angle) aproches the length of it projection on Y axis, which is sine of that angle.
For tan, because tanx = sinx/cosx, you know that for small x cosx aproaches 1, so tanx aproaches sinx/1, so tanx aproximately = sinx. 



#5
Dec1512, 05:40 PM

P: 106

Okay but how did they get that angular magnification is height of image / height of object?




#6
Dec1512, 06:22 PM

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It's harder to do the same thing with a microscope but the same principle applies. 



#7
Dec1512, 09:06 PM

P: 106

But when looking at angular magnification other factors like distance from the eye come into play as well, and that formula does not take them into account. For example, if the object and image are of same size, then according to that formula m=1 but if the image is farther away it is actually diminished as it will subtend a smaller angle at our eyes.




#8
Dec1612, 12:30 AM

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#9
Dec1612, 12:34 AM

P: 106

What about multiple reflections in two plane mirrors? There the images are the same size but their visual angle is less.




#10
Dec1612, 12:59 AM

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#11
Dec1612, 02:05 AM

P: 106

But it appears to be bigger and that's what angular magnification is. Just like trees farther away are smaller, or railway tracks converge at the horizon.




#13
Dec1612, 02:10 AM

P: 106

Because it's closer to you. Just like the trees right? The size of the image on the retina is bigger.




#14
Dec1612, 02:25 AM

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#15
Dec1612, 05:22 AM

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P: 11,352

When you try to assess 'size' in an objective way, your brain has a real difficulty. We still see Jumbo Jets lumbering at a leisurely speed across the skies of London because our brain tells us that something 'that big' couldn't be up there. So we 'see' a smaller aircraft going much slower than the real speed and assume it's only just above the roof tops.
There are many similar illusions. The only way to describe 'magnification' reliably has to be to describe the 'angle subtended' by object and image  as if the two were side by side at the same distance from us. The image we see in a telescope of microscope is only virtual, in any case and we can place it where we like by adjusting the eyepiece. This, we have to talk of angular size and not real size  the image of a flea in a microscope, adjusted to be 1km away, would be a hundred metres long  (logically) but it is a meaningless to talk in terms of length. @risch: Talking about the size of the image on your actual retina is a good approach because that implies angular size  the most objective description. Your thumb, at arms length really is the same size as the Sun (in those terms)  fair enough. 


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