a ray of light falls normally on an interface of two transparent media what is the angle of refraction?
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By snells law:\n\n(n1) Sin(Angle Of Incidence) = (n2) Sin(Angle Of Refraction)\n\nPut Angle of Incidence to be 0.\nYou will find the Angle of Refraction to be zero too.
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Whenever a ray enters from glass medium to air medium, the light is reflected as well as refracted. That is both reflection and refraction take place.\n\nWhen the angle of incidence (in glass) is zero, the angle of refraction (in air) is also zero. This is called normal incidence and there is no reflection inside the glass.\n\nWhen the angle of incidence is increased from zero, there is both reflection and refraction. For refection the angle of incidence is equal to the angle of reflection and for refraction the angle of refraction (in air) is greater than the angle of incidence (in glass).\n\nWhen the angle of incidence equals the critical angle there is both reflection and refraction. For refection the angle of incidence is equal to the angle of reflection and for refraction the angle of refraction (in air) equals 90 degree.\n\nWhen the angle of incidence exceeds the critical angle, then there is no refraction and there is only reflection and the angle of incidence is equal to angle of reflection.\n\nThe ray not only is reflected internally but also is totally reflected, i.e., there is no refraction.  That is there is TOTAL internal reflection.
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Total internal reflection. A diamond has a high index of refraction - about 2.418 - one of the highest among natural gemstones. Air, by contrast, has an index of refraction about 1. When a light ray enters one of the surfaces or facets of a diamond at any angle except directly perpendicular ("normal") to the surface, it is refracted (bent) toward the perpendicular (normal) to the surface.  The light ray continues to travel inside the diamond until it reaches another edge.\n\nHere is where it gets interesting.  If the angle that the light ray makes with that surface is large enough, the light ray cannot refract - bend enough - to exit the diamond. Instead, it is "totally internally reflected" and that light passes through the diamond to another edge (facet).  There, if the diamond is properly cut, it will exit the diamond, giving the diamond a sparkle as many rays of light entering from other directions are also totally internally reflected. \n\nOver the years, gem cutters have learned the proper angles that will maximize the total internal reflection - that is, trap as much light from as many different directions as possible and cause it to exit the diamond through other facets.
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Visible light is the most familiar form of electromagnetic waves.  Light is produced by the rearrangement of electrons in atoms or molecules.\n\nNewton's laws of reflection and refraction treats light as a particle:\n\nLaw of Reflection:  When light is reflected from a surface, the angle of incidence EQUALS the angle of reflection.  The angles are measured from the normal (perpendicular) to the surface.\n\nLaw of Refraction: Deals with the  way light changes its direction of propagation when it passes from one medium to another.  The light beam transmitted into the second medium is said to be refracted. \n\nThe angle of refraction bends toward the normal if a light beam passes from air into glass (the angle of reflection is less than the angle of incidence).   If the angle of reflection is greater than angle of incidence,  light is bent away from the normal as when it passes from glass to air.
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simple  for smooth surfaces, the angle of incidence is equal to the angle of reflection\nso the angle of reflection is going to be the same angle mentioned in the question
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prism. \n\n(polarizer - phase shift; lens - angle of relection,refraction; analyzer - phase)
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The answers are:\na)35deg\nb)50deg\n\nThings to remember: The angle of incidence is the angle between the beam of light and the line drawn perpendicular to the mirror, NOT between the beam of light and the surface of the mirror. Also, angle of incidence = angle of reflection, so, for part a), the beam reflects towards M2 at an angle of 35deg. Form a triangle with the two intersecting mirrors and this reflected beam (internal angles in a triangle add to 180 deg). You will find that the angle of incidence onto M2 is also 35deg.\n\nFor part b): the angle of reflection from M1 is 60deg, and again forming a triangle with the intersection of M1 and M2, you will find that the angle of incidence for M2 is 50deg.
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The index of refraction of a substance is equal to the ratio of the speed of light in a vacuum to the speed of light in the particular medium.\n\nn = c / v\n\nIf the ratio of c to v is 3 to 1.2, the the index of refraction must be 2.5\n\nn = 3/1.2 = 2.5
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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
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First realize that since the beam is passing through something, reflection has nothing to do with it.  The only possible answer is the one that mentions refraction.
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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.
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The refractive index will be (sin50/Sin 35)Please put the values yourself as I donot have a scientific calculator
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Atmospheric refraction is the deviation of light or other electromagnetic wave from a straight line as it passes through the atmosphere due to the variation in air density as a function of altitude.\n\nAtmospheric refraction near the ground produces mirages and can make distant objects appear to shimmer or ripple.
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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.
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Can you be more specific? Every material has it's own index of refraction....and none of them work out to nice even numbers. (well, hardly any of them). Is this all the questions says? I looked around on some websites, none of which made this question any clearer. Good luck!
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when a light ray passes from a denser medium to a rarer medium, the refracted ray bends away from the normal and the angle of refraction is greater than the angle of incidence.\nso, when the angle of incidence is continuously increasing the angle of refraction also increases in such a way that agile of refraction is always greater than angle of incidence, till for a particular value of Angle of incidence , the corresponding angle of refraction becomes 90 and the refracted ray grazes along the surface.\nthe angle of incidence in denser medium for which corresponding angle of refraction is 90 in the rarer medium is called the critical angle denoted by C.
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Have a look at the reference for an explanation of Bragg's Law\n\nIf the surface plane is plane 0, the next one plane 1 etc.\n\nThe path difference (PD) between plane 0 and plane 1 is 2d sin theta = 1.01 lambda\n\nThe PD between plane 0 and plane a is 2ad sin theta = 1.01*a*lambda\n\nCancellation occurs when the waves are out of phase by half a wavelength, i.e. the path difference is lambda times [an integer (whole number) plus 0.5]\n\nSo we’re looking for a vale of a where 1.01*a = integer + 0.5\n\nSuitable values are plane 50 (PD = 50.5), plane 150 (PD = 151.5), plane 250, 350 etc.\n\nTip: Don't get into X-ray crystallography, it'll drive you nuts!