Close up lens +1, +2 and +4. Hoiw magnification works.?
361687
Use the lense equation.\n\n1/f = 1/u + 1/v\n\nF is the focal length of the lense (1/f is the power of the lens as in your question)\nU & V are the near and far sighted distances.\n\nFor close up photography assume v= infinity (1/v = 0)\nU is the distance from your lens of the object being photographed (in metres).\n\nIe if you are photgraphic 20cm away, u=0.2. 1/u = 5 = the power of the lense you want.
534267
You won't find them with a -2.75 BC (base curve).  -2.75 is the power.  The base curve will be a number like 8.3 or 8.6.  If you're confused about your prescription, you should talk to your eye doctor.
1149797
The first one is (-5, 3). I don't remember how to do the others.
166306
You need to know the focal length of the eyepiece as well.  The magnification is the ratio of these focal lengths; e.g., 48 if you have a 1 cm eyepiece.  The apparent distance is the actual distance divided by the magnification.
925522
10x50 is 10x magnification and a 50mm objective lens\n20x70 is 20x magnification and a 70mm objective lens\n A larger objective lens will allow more light in for clearer viewing, especially in low light conditions, and allow a larger field of view at any given distance.
918247
what it means is this 4 plus negative .01 and with the other it's 5 plus negative .04 its simple. if you know what it means and if you took notes look at them. it may help a little
774120
Sure,\nYou are adding two vectors:\n\nV1 has magnitue 0.45 m/s and in the W direction (head of vector faces W) \n\nThe 2nd vector is has magnitude 0.35 m/s and 30 degrees E of N (or 60 degrees from +ve x-axis) and its tail is added to the head of the first). So the angle betwen the two vectors is 60 degrees.\n\nUse the SAS rule for finding the magnitude of the resultant vector\n\nc = sqrt(a^2+b^2 -2ab cos(theta))\n\nVr = sqrt((0.45)^2 +(0.35)^2 - 2*0.45*0.35 cos60)\n   = 0.409 m/sec \n\nThe direction of Vr is from the tail of the first vector to the head of the 2nd.\n\nTo determine the angle, A, bewteen Vr and the first vector use the sine law.\n\n0.409/sin60 = 0.35/sinA\n\nSinA = 0.74 and A= 47.8 degrees\n\nSo the direction of the craft is 90-47.8 = 42.2 degrees W of N
955805
That simplifies to 3[2x-1]>12\n\n[2x-1]>4\n\n2x-1>4    2x-1<-4\n\nx>5/2  or x<-3/2\n\nWhen you have absolute values, you have to look at both ends of the spectrum
757540
The equation which relates the object distance (Do), the image distance (Di), and the focal length (f) of the mirror is\n\n1/Do + 1/Di = 1/f\n\nSolving for the image distance\n\nDi = 1/(1/f - 1/Do) = f*Do/(Do - f)\n\nFor this case, the image will be minified and inverted. FYI, the magnification is given by M = -Di/Do, where a negative value indicates and inverted image. These results can be verified through ray tracing.
160744
I suspect that the focal length would change sign.  That is, a convex lens would now act like a concave lens and vise versa.
635627
How about a polarized lens
1116225
simply multiply it with 0.4
1122353
Check out PhotoImpression 4
793064
10x the objective lens.\n\n;-D That is why they put that little 'x' in there.
1136934
flock or something, but thenks for the +2
1086553
Way back in the olden days, I used to take my SLR with 2 lenses, 28-50 for wide shots and 75-200 with a 2x coupler for close ups. Never had a problem, but then again there was no terrorists problems back then. I still have some nice shots of Mike Schmidt, Greg Luzinski and Steve Carlton somewhere at my mother's house. I think as long as you let them search your bag, you're okay.
45435
commmon one is Convex lens\n(with different angles, to magnify upto diff levels)\nbut for higher levels, it depends upto the magnifying object and distance also