calculate index of refraction of the unkonwn?
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
926057
The above solution has to be repeated again on the inside of \nthe prism just before the light comes out the other side.\n\nThe light gets refracted twice, one on contact with the prism, and\nagain before exiting the prism.  So you have to calculate the\nangle of incidence on the inside of the prism given the apex\nangle of 46 and the refracted angle of 26.36. A geometry\nproblem. Then repeat above calculation.
277776
when looking into a bowl of water. Air has a refractive index of about 1.0003, and water has a refractive index of about 1.33. If a person looks at a straight object, such as a pencil, which is placed at a slant, partially in the water, the object appears to bend at the water's surface. This is due to the bending of light rays as they move from the water to the air. Once the rays reach the eye, the eye traces them back as straight lines (lines of sight). The lines of sight intersect at a higher position than where the actual rays originated. This causes the pencil to appear higher and the water to appear shallower than it really is. The depth that the water appears to be when viewed from above is known as the apparent depth.
440843
numerical recipes in C.  Amazon has it.
419817
deviation= (refractive index -1)*angle of prism\n\nif prism angle is small...\nand\ndependence of rf with wavelength is given by cauchy's formula\n\nrf= A/wavelength....  A is constant\n\nso dependence (deviation) proportional to [(1/wavelength) -1)]
420514
go to this site and look at the image:\nhttp://en.wikipedia.org/wiki/Concave_lens\n\nmake a similar experiment where 3 beams come in parallel and diverge after they come out. then imagine that the lens wasnt there, and the diverging beams all came from the same point, called the focal point. the distance from the lens to this point, which will be negative since the point is imaginary and behind the lens, is the focal length, measured in meters. if you take the reciprocal, 1 / focal length, you will get the lens power, measured in diopters
1136420
I think about the refraction of light.
727712
Speed of Light in Air = 2.99792458 X 10^8m/s\n\nTo get the refractive index,take the speed of light in air divide by the speed of the in another medium.\n\nHence is (2.99792458 X 10^8)/(2.5 X 10^8) = 1.2(approximate)
63098
Sqrt(Pi/2) * Fresnel [Sqrt(2/Pi)*x]
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).
362195
Wavelength= 1/wavenumber\n\nWavelength= 1/1700cm-1\nWavelength= 5.882 x 10^-4 cm\nWavelength= 5882nm
774526
On a smooth surface, the angle of reflection will be the same as the angle of incidence.\n\nTherefore, the answer is (B) 40 degrees.
852084
http://www.stevenseow.com/tutorials/QuantifyingTheBlindSpot.html
42425
The index if refraction (n) of a medium is the ratio of the speed of light in a vacuum to the speed of light in that medium.\n\nn=c/v \nc = speed of light in a vacuum\nv = speed of light in the medium\n\nn is almost always greater than one, as in most cases an electromagnetic wave will travel faster in the vacuum.
485679
refractive index = 1 / sin of critical angle\nrefractive index is 1 / sin30\n1/1/2\n=2
519599
The index of refraction of water is 1.33.  The refraction of light from the air to the water creates an apparent magnification of objects below the surface.  This magnification gives the impression that things are closer than they are and that the water looks shallower.
359061
As light passes through more and more optically dense media, the speed at which light wave front propigates slows down and is bent.\n\nThe amount which the light ray is bent depends on the indicies of refraction of the two media (the one the light is leaving and the one it is entering).  Using Snell's law, we can calculate the angle as,\nn_1 * sin(angle 1) = n_2 * sin (angle 2)\nwhere n is the index of refraction and angle is the angle between the light ray and the normal line on the medium's surface.\n\nKnowing the index of refraction, we can calculate the speed of light in a particular medium.\nv = c / n\nwhere v is the speed of light through the medium, c is the speed of light in a vacuum, and n is the index of refraction.\n\nAs light passes from one medium to another, the speed at which it propigates slows down and the wavelength changes, however, the frequency of the light remains constant.