Can someone tell me without being too scienctific on how diverging lenses work?
485221
Yes.  A diverging lens "demagnifies"  It makes objects smaller rather than larger.  These are used for people who can't see far off.
52768
B.  A convex mirror will make light diverge, and no mirror will *alwats* give a inverted image, because it depends whether you're closer than the focal point for a concave lens.\n\nThink about those little round mirrors on buses and huge trucks, and how small the image is.  Those mirrors are convex.
294491
My son has a cheap red laser pointer (well under $10)\nwhich has a few "lenses" that he can screw on the\nfront to make different patterns.\n\nI don't know if these things are strictly "lenses",\nbut they distort/focus/shape the laser dot to\nlook like a star, a line, or a couple of other\npatterns.\n\nIn any case, one causes the laser pointer to produce a\n"line" rather than a "dot", and that seems like it would\nsatisfy some of your needs.  The width isn't readily\nadjustable, but the the length would be (by covering\npart of the "line" just past the light source).
928848
What's not clear is whether the image is a virtual or real one.\n\nFirst, consider a quick graphical, geometrical ray trace of a focusing \nmirror (let's assume it's spherical so that the analysis is simple).  I \nusually do this by drawing the reflector and locating its center of \ncurvature and focal point (which is just half the distance from the middle \nof the sphere to the center of curvature).  To look at image formation, I \ndraw a large number "1" where the object should be -- in this case, \noutside the focal point.  I then draw two rays from the top of the "1" -- \none parallel to the line from the center of the sphere throught the radius \nof curvature and one through the radius of curvature.  The first ray must, \nafter reflection, go through the focal point (because it acts as a \nparallel ray from infinity, which are all focused at the focal distance \nfrom any optic).  The second must just come back upon itself, because it \nis perpendicular to the reflector surface.  The intersection of the two \nrays is where the head of the image "1" is located.  In this case, it's \noutside the focal point.  I then do the same thing from the bottom of \nthe "1", and then draw the orientation of the imaged "1"...  It's clearly \ninverted and reduced in size and more than 1 focal length away from the \nmirror.  This is clearly the real image you referred to in your question.  \n\nWhere's the virtual image?  It's not there!  What you're observing is the \nreal image.\n\nTo understand this, consider what a virtual image is and how it's formed \nby an example.  A virtual image does NOT result from the convergence of \nray bundles at any real point in space... A virtual image is observed when \na situation arises rays from a single point appear to emerge from a point \nin space, but which doesn't correspond to any real point.  A good way to \nsee this is to turn the spoon around and look at the convex surface.  You \nsee yourself right side up!  If you do the ray trace using the method I \noutline above, you find an apparent convergence of rays behind the mirror \nwhich produces an upright image, which is just what you see in this case.  \nBut, since the spoon is silver, it's pretty apparent that there can't be \nany rays actually focused behind it -- so it's a virtual image.\n\nAn interesting thing occurs, however, when you flip the spoon back to the \noriginal configuration and move the object inside the focal point.  The \nimage you see is no longer inverted! And it's big!  If you do the ray \ntrace you find that the convergence of the rays which form the image is \nnow in back of the spoon, so that the image has shifted to an upright \nvirtual image which is magnified.  Try it with one of your spoons -- it \nreally works!\n\nAnd even more interestingly, if the object were placed exactly at the \nfocal point, all hell would break loose and what you would see would be a \ngiant blob filling the spoon that looks like nothing but a big source of \nlight.  In this case, there is neither a virtual nor a real image formed \nand the mirror/spoon is acting like a collimator -- turning all rays from \nthe object into parallel light streaming to infinity (roughly).\n\nI hope this helps!
984480
buy two convex lens, both should have different focal lengths and it is advisable that you should buy lenses of high focal lengths. and a tube is also required.the length of the tube should be equal to the sum of the focal lengths of the two lenses.\nNow, make the bigger lens as the eyepiece and the smaller one as objective. your telescope is ready to wander through the universe.
487323
ok, well think about what the resolving power means, first.  \n\nit says that if you take 2 point-like objects, and bring them closer and closer together, then your eye (or some optical instrument) will have a lower limit in terms of seeing them as 2 individual objects.  at some distance apart from one another, they will simply blur together and look like a single object or blobby thing.\n\nnow obviously, if the 2 objects are closer to your eye, then you can see them better, so the "distance between objects" is maybe not a very good way of determining your resolving power, because it depends on where the objects are.\n\na distance-independent way of stating the resolving power is in terms of the ANGLE between the objects.  this is the angle formed by the line that goes from object 1 to your eye, and the line that goes from object 2 to your eye.  for any given eye, there is some angular separation between 2 objects below which distinct objects cannot be resolved.  this angle is called the resolving power.  (radians are just a unit of angle, like degrees.  there are 2 * pi = 6.28 radians all the way around a circle.) \n\nso, if you think about it, someone with sharper vision can resolve 2 point-like objects that are closer together than can someone with poor vision.  this corresponds to a *smaller* angle because they're closer together.  so a small value for resolving power corresponds to sharp eyesight.\n\nas for optical or visual acuity, that is simply the reciprocal of the resolving power.  that means that a *larger* value means sharper eyesight.  in a medical context, visual acuity is usually expressed as something like 20/10 (which equals 2) for very good eyesight or 20/100 (which equals 0.2) for poor eyesight.  as you can see, the larger the value, the better the eyesight.\n\nhope this helps!\n\nps - have a look at this website if you need more info:\nhttp://www.tedmontgomery.com/the_eye/acuity.html
1179980
Just as with any other kind of lens, the shape of the lens will be one of the parameters that affects the focal length, but I’ll assume you will use the same-shaped container to look at the differences between different liquids.  (The material that the container is made of will also affect the focal length, but I’ll also assume that’s constant, too.)\n\nIn this case, there are really only three things that matter, and if you use a single color light source (e.g., a red laser pointer), only the first matters.  Even if you don’t, the second parameters listed below are normally of much less importance than the first.\n\n1.The refractive index of the liquid\n2.The chromatic dispersion of the liquid\n3.The wavelength (color) of the light being sent through the lens.\n\nThe primary parameter is the refractive index; which quantifies the extent to which a light ray is refracted (“bent”) as it crosses the interface with the liquid and whatever is on the other side of the liquid (i.e. the container material).  For a given shaped lens, a higher refractive index liquid will cause more refraction, and hence result in a “stronger” lens.\n\nThe chromatic dispersion tells you how much the refractive index varies as a function of the wavelength of the light.  All real materials have some dispersion, and because the refractive index for different colors of light are slightly different, the focal length and focal point of any real lens is slightly different for different colors.  (Manufacturers of things like high-quality camera lenses take great pains to minimize the dispersion of their lenses by using multiple optical elements (individual lenses) made of different materials.)\n\nFinally, because the index of refraction varies as a function of wavelength, the focal length will also depend on the wavelength of the light you use.\n\nIf we ignore the last two effects (e.g., let’s say you use a laser pointer as the light source, we can concentrate on how you might vary the refractive index of liquid you use.  First, different liquids have different refractive indices because they have different chemical compositions and molecular makeups.  Materials have characteristic refractive indices because of the way light interacts with the atoms and molecules making up the liquids.  Different molecules interact with light in different ways, giving rise to differences in refractive index.  \n\nYour first experiment might then look at different composition liquids (oil, water, alcohol, water containing the same concentration of different dissolved substances, water containing different concentrations of the same substance, etc.).\n\nSecond, to a much lesser extent, the refractive index of a material depends on temperature (mostly because the density of materials depends on temperature).  You could try looking at the effect of heating the same fluid to different temperatures and seeing if that makes a measurable difference in the focal length.  Different fluids will have different temperature dependences.
436655
Hello?! You see people every day who are walking around with examples of Snell's law at work -- eyeglasses!\n\nSnell's law applies to optics including the lenses in eyeglasses, telescope lenses, overhead projectors, video projectors.  Then there are prisms such in mobiles.  And the sparkle in diamond rings.  And why when you see a fish in a pond, the fish isn't actually where it appears to be.
497011
It works by having little mirror slices at different angles. As light is reflected off the orb, the little slices separates the light into several colours briefly and the process is repeated.
303025
A convex mirror diverges light outwards. When the image is observed, the image appears to come from a point behind the mirror. This is a virtual image because the light never actually comes from this point, and it is not possible to focus the image onto a sheet of paper.
1109140
You can really only see images that are focused onto a small little center on the back of your eyeball.  When your lens becomes distorted, the light is reflected onto the wrong area.  Normally all light converges on this area and you see 1 image.  The images you are seeing are overlapping and appear blurry to you.
1022562
A Fresnel lens works just the same as a standard convex lens (or any positive lens).  The only difference is that a Fresnel lens is flat!\n\nThe magnification is related to the focal length of the lens.\n\nHere is a good explanation for calculating the magnification...\nhttp://en.wikipedia.org/wiki/Magnification
152081
well, each ocular lens usually has the same magnification, otherwise you get a headache as you will see the resulting image in different sizes at each eye.\nAnd each eye sees only through one ocular lens, so how could they possibly affect each other?\nYou only multiply the effect of lenses which are in series (one behind each other).
913177
Yes there are, and the best (imho) explanations in plain, non-technical language are in a book by Richard Feynmann titled  "QED The strange theory of light and matter".\nWhen you first read what actually happens when light bounces off a mirror it's mind blowing. There are equally good explanations of why a simple lens works, etc.
949345
It's all about refraction ( a fancy term which means bending light)... The lenses in the microscope bend the light before it enters your eye so that it looks as though it comes from a much bigger object. Draw a horizontal line and put a dot on the left of the line to indicate your eye. On the right of the line put a vertical stick (so its base is on the line).  Now join the top of the line to your eye. That's a light ray entering your eye from the top of the stick. Now make the stick much bigger. and do the same thing. Notice that as the stick gets bigger the angle between the horizontal line and the light ray from the top of the stick and your eye gets bigger too. If you can change the angle as light enters your eye things will look biger. That's what the lens does.
417004
there are little muscles above and below the lens of the eye, which, in profile, is shaped like an oval or elipse.  When those muscles are relaxed or flexed, the oval is either stretched or becomes shorter, changing the shape of the lens, thus manipulating lightrays (reflecting off of the thing we are "seeing") traveling through the lens, so that these rays (depending on the distance you are from what you're looking at and what you're doing with those muscles) are more focused or less focused in the retina.  If the light rays aren't focused correctly back there, then you've got a blurry image.  At least, that's what I THINK is going on if I remember correctly from a class I took last year....  So, in short, it's our ability to manipulate our eyes' lenses with those little muscles that makes us able to blur (or focus) our vision... good luck.
406370
More light is needed as the magnification is increased since the size of the hole of the objective lens decreases. Try it!