What does 4X12X40 mean in rifle scopes? 40 is the milimetres of the lens buta what of the rest?
360890
I don't know a thing about guns, but this page seems to suggest that it's 4-12X zoom.
302378
1dm=10cm (deci means 10)\n400dm=4000cm\n400cm<4000cm\n400cm<400dm
829483
for the size of the lens i would go back to the lens maker formula it is:\npower of lens = (n(lens) - n(0)) * (1/R1 - 1/R2)\nwhere n(lens) is refractive index of lens material\nn(0) is refractive index of the medium in which lens is kept\nR1 and R2 are surface radius of the two sides of the lens.\nso you can see from the equation to increase power of the lens we can either increase refractive index of the lens (assuming we are seeing it in air) or reduce the radius of curvature of the lens sides. so ideally as power increses the radius of the lens decreases and so the size of the lens also reduces. you can also check out gullstrands equation for further reference of thicker lenses. hope this helps
882813
according to gun traders guide 350.00 quality brand  of scope would add maybe 125 150 to total ..if cheap scope minimal amount...made from 1948-62..also local market and season will affect price
44487
Something is wrong with this question. I see it is 2 months old and would write directly if you allow email.\nI think the question should be about the object not the image. If the object is at the focal length of the lens, then the rays are parallel and easy to see with the eyes. The only image is on the inside of the eye.\nIt doesn't make sense to have length anywhere.
763008
A focal length, in a lens or mirror, is the distance from the lens to the focal point (the point where all rays converge from being directed by the lens).
439488
1 mil = 1/1000 of an inch.\n\n12 mils = 0.012 inches.
409559
The light gathering power and resolution are determined by the size of the objective lens, but not of its focal length. The ease of going to higher powers is determined by focal length.
363353
It's probably the water causing refraction of the scope.  You'll see that at higher magnifications.  I think it's the same principle as a snowflake.
786859
Well, its smaller than a 12 or 16 gauge, but larger than a .410, so it just depends on what you're comparing it to.  I prefer a 12 for most of my shotgun hunting, but I know plenty of people who love bird hunting with a 20.  I suppose it just depends on what you want to use it for, but a 20 gauge has plenty of power to get the job done in most cases.
27778
5.56 x 45\n\n.223 rem
1134
It depends.  Just about any binoculars provide good view.  Make certain you stay away from very strong magnifications, otherwise you will need a tripod to keep them steady and may get dizzy without it.  The larger magnifications also limit your field of view.  \n\nI have some 8x42 Nikons that are great.  \n\nJust in case you are not aware:  the first number given in the binocular is the magnification.  Stick with 7 or 8.  The 10s you will need the stabilization electronics or a mono/tripod.  \n\nThe second number provides information on how much light comes through the lens.  \n\nFor example...if you were to compare the same brand and model with the exception of the number difference: 8x25 vs 8x42:  \n\nThe 8x25 will provide the same size image but the 8x42 will provide a brighter image.   If you are thinking about purchasing online...before you do find a camera store or a sporting goods store to see how they feel and get a better feel for what you really want.  Hope that helps...happy star viewing!\n\nAlso another possible source for help would be to see if there is a local star party group in your area.  Drop by one night; it can be fun to meet some experienced viewers with telescopes already setup as well as some people who have the really large expensive scopes which you may not be able to afford.
1095598
For part 1, if the cell's diameter was 1/15 of the total field, and the field was 450 microns across, then the cell's actual diameter was 450 microns/15 = 30 microns = 3 * 10^-5 meters.\n\nThe diameter of this cell in the drawing is 0.5 cm = 5*10^-3 meters.  This drawing therefore has a magnification of (5*10^-3)/(3*10^-5) = 166.6666666...times.\n\nFor part 2, The magnification is given by (size in image)/(actual size), so we have that 45000 = (1.5 mm)/a, where "a" is the actual thickness.  Solving for a, we get a = 1.5mm/45000 = 3.333*10^-8 meters = 33.3 nanometers = 0.0333 microns.\n\n-----------------------\n\nAs for the question posed in the "additional details", when the magnification of a microscope is increased, the diameter of the field being imaged decreases (choice "A").
780266
Bound's hubby here:\n\nThe height of your scope rings is directly dependent upon the diameter of your scope's objective lens and your rifle barrel's contour!\n\nOne thing you need to keep in mind, is that the ring height you select will have an effect on how you place your face on the cheek piece! The taller the ring, the higher the cheek piece you be in your stock! For the 32 mm objective, I would use the low rings ... if and when you move to a 42mm objective, then get the taller rings!\n\nGood luck!
537574
no, the focal lenght is the property of the mirror, not of the medium in what is situated.\nthe focal lenght = mirror radius / 2
978537
Mostly clarity, a well focused image across the whole field of view with well focussed crosshairs. Has he suggested a magnification? Does he want variable power? For fixed power I mostly go with 6x for hunting, higher magnification seems to make it slow to get the sight picture right and acquire the target. For variables something in the 2-7 to 4-12 range should do, 3-9 is the most popular (or was). I'd be interested to hear any alternative views on those. \n\nWaterproof and fogproof too, don't want to forget those. You need to ask if he want's glossy, matt or stainless too, it would be a shame to get him the wrong thing, if he's bothered.\n\nFor best brightness the objective diameter should be 7x the magnification, so a 4x28 gives a 7mm exit pupil, which means that the light it gives out can all be taken in through the pupil of an eye. If you have a 10x40 then the exit pupil is 4mm and only about half the pupil diameter of the human eye receives light. That's why we find funny numbers like 8x56, 6x42, 4x28 and also why a lot of sailing binoculars are 7x50. This is at the esoteric end of things though, focussed is much more important.\n\nCabelas is another Leupold stockist.
830197
You multiply the two numbers, so 58 x 10 = 580 total magnification.\n\nGood luck!!