How do you calculate Numerical Aperture (NA) for a lens?
455316
the numerical aperture of a lens is defined as \nNA = n sin theta \n\nwhere n is the refractive index of the medium that surrounds the lens (n=1 for air, 1.33 for water) and theta is the half-angle of the cone of light that can enter the lens.  obviously this will depend to some extent on the specific geometry of the lens' surroundings.   essentially the numerical aperture relates the focal length of the lens to its aperture.   this is sometimes called the f-number or f/stop which is defined as f = 1/(2 * NA).
643486
Assuming you mean the index of refraction of a totally saturated salt water solution, you can use that plus the index of refraction of water to find the concentration of the salt solution through proportionality. The more salt you add, the closer the index of refraction will get to that of the saturated salt water solution. Put mathematically (and let's say for fun the index of refraction of salt water is 1.50 but I have no idea what it is offhand. Water, it's 1.33.)\n1.33x + 1.50y = (whatever you found to be your index for your solution)\nx + y = 1\nSolving for y will give you the concentration as a fraction of the concentration of totally saturated salt water. Hope that helps!
925293
c = lambda x nu\n\nc is the speed of light 3.0 x 10^8 m/s\n\nlambda is the wavelength\n\nnu is the frequency
894614
PV=nRT\nP= (nRT)/V  (1st eq)\n\nn,R,T u know.\n\nV >>>\nDensity = mass/ volume\n2.3 = .00446 *10^3/Volume\n\nrearrange and ull get volume.............and then put that into the first equation.............and voila
370398
O.K.\n\nUse the substitution:\n\ny = uv , x = u/v\n==>\n(1+u^2)(u/v)(udv + vdu) + (1-u^2)uv(du/v - udv/v^2) = 0\n==>\n(1+u^2)(udv + vdu) + (1-u^2)(vdu-udv) = 0\n==>\n2vdu + u^2(2udv) = 0\n==>\ndu/u^3 + dv/v = 0\n\nIntegrating gives:\n\nv = (constant) e^(1/(2u^2))\n==>\nv^2 = A e^(1/u^2)\n\nReverse substituting x,y gives the answer as an implicit function:\n\ny/x = A exp(1/xy)
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.
785008
_______I_____I_____()_____I__________________\n              o          f                      f \nas the law of convex lense is that is a ray is passing from optical center<that is the mid of the convex lens >then there will be no change in its direction and if another ray is paralell to principal axis <that is the long straight line>then the ray will pass from principle focus<that is f>then keep on drawing the rays until they converge together at a point that will be ur image ......after taht u can use the formula to calculate it or just measure the distances\n1/image distance+1/object distance=1/focal length\nu can change it to\n1/image distance=1/focal length-1/object distance\n                            =1/3-1/5\n                                   5-3\n                             =_____________\n                                   15\n                             =2\n                               __________\n                                   15\n now we will flip it to\n      15\n_________\n        2\nand then we will get 7.5 answer\ni hopedi helped u i m studying this too right now in 0 levels i have a lot of h.w about it today
5733
7x50 WP-XL \n \n         \n \nMagnification     7 \n \nObj. Lens Dia. (mm)     50 \n \nEye Relief (mm)     18 \n \nApparent Field of View     7° 30' \n \nField of View at 1000m     131 \n \nExit Pupil Dia. (mm)     7.1 \n \nRelative Brightness     51 \n \nInterpupillary Distance     56~72 \n \nLength (mm)     184 \n \nMax Width (mm)     194 \n \nWeight (lbs.)     1.8 \n \nRubber Coated     Yes \n\nDo you need to know how to use them?
247778
http://focus.ti.com/lit/ml/slup117/slup117.pdf
1068729
i think n=6..\nyes this can be done..if you know the formula..\nof volume and surface arae of a hypersphere and also the form of gamma fucntion..\nhope you can do it yourself..i saw these two links and based on the formula given for volume and surface ares and the gamma fucntion expression its evident that n=6.\nHope you can figure it out yourself.\nvolume= pi^3/6 (radius =1 here)
611398
The ratios of the indices of refraction will be the same as the ratio of the sin of the angles.\n\nsin(32)/sin(27) = x/1.61
541329
try www.lenscrafters.com or .net
583825
take above as :-\nm= -v/u = h'/h = 2h/h = 2\n-v/u = 2, v = -2u\ntake,\n1/f = 1/v + 1/u\n1/-50cm = 1/-2u + 1/u\n(-50) because focal length of concave mirror is always negative.\n\n\nnow, solve it get value of 'u'& you will get ur object distance.\nit must be less than 0.5 m as the image is erect & enlarged.
241224
As best I recall, the focal length is equal to 2*the Radius of Curvature; 1/f = 1/o+1/i\n\nM= -{i/[o]}; thus, f=54; the image distance is 3 * the object distance; 3o + o = 54 cm; 4*[o] =54 ; [o]=13.5; the other object distance ? I can only find one; the sign on the image is distance is +; Magnification is - so it is "inverted"; or if o=-13.5, the second answer, I THINK, is found; with a +3 magnification; upright
950045
Correct formula should be as follows:\n\nE = 1/2*I*w^2
155010
You need an achromat, which is a compound lens (doublet) designed to have the same focal length at multiple wavelengths.  Take a look at this on from CVI lasers:http://www.cvilaser.com/Catalog/Pages/Template1.aspx?pcid=196&filter=0
873982
c to f:\nn*1.8; add 32\n\nf to c:\nn-32; multiply by 0.5555