Find the critical angle of a piece of glass that has an index of refraction of 1.45.?
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READ THIS CAREFULLY\n\nCritical angle is that angle of incedence in denser medium for which the angle of refraction is 90 degrees\n\nhence\nacc to snells law\n\nsin i/sinr = 1/refractive index\n\nhence\nsin i/1 = 1/1.45 [sin 90=1]\n           =0.69\ni= sin^-1[0.69][pronounced sin inverse of 0.69]\n =43.6
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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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the external angle\n\ni hope that this helps
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Yes...\n\nThe refraction indecces\n\nn =c/v, but c = wave lenght * frequency
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If you are given the arc length of s,\n\nThe angle may be found by\n\ns/(2*PI*r) * 360 degrees.
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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!
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easy. For a basic refractor just reflects light through lense then bends by way of mirror into eyepiece which also has lense. Focusing lense. Hope helps.
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assuming "perfect" polarization:\n\n50% through the first.   Then use trig to calculate the E field amplitude reduction caused by the 42deg  polariztion.
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Because of the difference in the index of refraction, with everything else equal, the glass should be brighter because it is less diffuse.\n\n\nhttp://www.findarticles.com/p/articles/mi_rlg4413/is_SLGG0003/ai_n15607430\n\nhttp://www.madsci.org/posts/archives/Jan2003/1042659817.Ph.r.html
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the angle between the incident ray and the perpendicular to the reflecting or refracting surface at the point of incidence
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http://hyperphysics.phy-astr.gsu.edu/hbase/electronic/posfe.html#c3\n\nThe link above may be of some help. Input the values you want or need.
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Refractive index depend uopn the material. Unfortunately u did not maintain the material. The value is sin i/sin r, where i=angle if incidence and r=angle of refraction. So refractive index cannot be zero, all other value can be possible for different material.
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wellllllllllllllllllll a nice and easy question with an extra easy answer\n\nuse the formula \n                \n             1/f=1/u - 1/v \n\nwhere according to sign convention f,u will be negative. since it is left to the lens v=-8.2cm
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find the area of the cross section.... then times the area by the length... that is of course if it's a prism
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Try: allaboutphysics.com I think that might help.
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Assume that the incident rays are incident normally on the flat side (one of the sides forming the right angle) of the prism. The side opposite to the top angle is the base. When any light is incident perpendicular onto a transparent medium, it is simply transmitted without undergoing any refraction. So, the ray incident on the flat surface normally (means perpendicularly) is transmitted to the slant face containing 65* angle. Construct a line perpendicular to this slant face at the point of incidence. You will find that the angle between this line and the light rays is 25*, also known as the Angle of Incidence. Using\nRefractive Index = (Sine of the Angle of Incidence/Sine of the Angle of Refraction), we can find the Angle of Refraction for each colour depending upon its Refractive Index. The Angle of Incidence is the same 25* for both the rays. The problem is that the Refractive Indices given are for light travelling from air medium into glass medium, but, here, the light rays are travelling from the inside of the glass medium to the outside air medium. So, reverse the Angles of Incidence and Refraction so that it looks as if the rays are travelling from the air medium into the glass medium. So, now, the <i is to be found out and <r = 25*\nU = (sin <i / sin <r), sin <i = (U x sin <r). Calculating this, <i for red = 44.62* and <i for violet = 45.857* These are the answers. <i is the Angle of Incidence and <r is the Angle of Refraction.\n\nStill doubts? Visit www.pcmhelp.netfirms.com
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_______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