00:01
All right, so for this question, he was asking us, what is the largest overall magnification? okay, so we know things we want to care about the value of magnification.
00:15
We can just simply put absolute value on m here.
00:19
M here stands for the magnification, which is equal to absolute value minus v over you, which is equal to absolute value v over you, okay? the image distance, u is the object distance.
00:35
So in this question, you're saying that each object lens is forming an image 120 millimeter beyond its second focal point.
00:43
Okay, so the image distance formed by the objective lens is much larger when compared to very focal lens.
00:50
And this only happens when the object distance is nearly equal to the focal lens of the lens, which means that we basically need to consider the object distance is equal to the focal lens.
01:03
So now we have such an equation with absolute value m equal to v over f.
01:10
So now let's start plugging all different focal lens and the image distance to find out different magnification number.
01:19
Okay? so well, just want to let you guys know image distance here is equal to the distance apart, let's say l, plus the focal lens.
01:31
Okay? and in the question you were saying that it's one 120 millimeter beyond.
01:37
So it's 120 millimeter plus the whatever focal lens you have.
01:46
So first, let's focus on the focal lens of 16 millimeter.
01:51
So we got m1 is equal to, you want to put absolute value, that is totally fine.
02:01
With a v which is 120 millimeter plus 16 millimeter over 16 millimeter and this will give us the values about 8 .5.
02:26
So for the focal lens with 4 millimeter it will have a magnification number of b over f which is 120 millimeter plus 4 millimeter over 4 millimeter and this will give us 31 for the magnification number and a magnification number for the focal lens of 1 .9 mm -upre 1 .9 millimeter will give us v which is 120 millimeter plus 1 .9 millimeter over the focal lens which is 1 .9 millimeter and this will give us about 64 .16.
03:31
Okay.
03:32
So now we know that comparing with all its focal lens, the magnification number is the largest when the focal lens is 1 .9 millimeter.
03:45
Okay, let me use it here.
03:47
So this is the largest.
03:53
Remember in a question we're saying that ips have been angular magnifications of five times and ten times each objective form, which means that in order to match the largest magnification number, we need to match the ips with 10 times magnification number with the focal length of 1 .9 millimeter, which will give us the magnification number 64 .16...