if an angle of incidence on aplane mirror is0 ,what the angle of refraction?
601884
mirror is designed to reflect the light and no refraction light .. from it .. so the reflection angel will be zero
400952
Use 1/s+1/s'=1/f  where f=R/2, s is object distance (in this case negative) where the image of the first mirror is the object of the second.  s' is the image distance you are solving for.  I'll leave it for you to work out the details of your homework, but this should get you started in the right direction.
489404
yes...but u must know the angle of incidence and angle of refraction...light up a candle and place a mirror opposite to it in a room at an angle so tat u can see the candle.then keep other mirror at a place where u can see the first mirror and the candle... continue the process till u get sufficient light. try to place mirrors in the ceiling to get better light.
440395
Refracted light rays bend toward the normal when the material being entered has a higher index of refraction then the material you are coming from.  Reflected light behaves according to total internal reflection, and bounces off the surface at the same angle that it approached at, much like skipping a stone off a lake's surface.  Check out snell's law on wikipedia for more.
571366
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
282836
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.
221019
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!
535452
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
116013
do u mean speed? no, the magnitude is stillt he same but the direction is changed if and only if the mirror has a smooth surface. cause if it is not. there would be many angle of reflection
696403
The outgoing intensity is found by applying the effective transmittance of each of the three polarizers, in order.  Assuming ideal polarizers:\n\n(I out) = (I in) * (50%) * (cos(24))^2 * (cos(45-24))^2\n\nso (I in) = (2.3 W/m^2) / (0.5) / (.834565) / (.871572) = 6.324032\n\nor about 6.324 W/m^2\n\nNote that this assumes the 45 degree angle is measured in the same rotation direction (e.g., both CCW) as the 24 degree angle.  If the 45 degree angle is measured in the other direction (e.g., CW rather than CCW), then the factor (cos(45-24))^2 should instead be (cos(45+24))^2, which would give the value of 42.918 W/m^2 for the intensity of the incident light.
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.
751077
Yes...\n\nThe refraction indecces\n\nn =c/v, but c = wave lenght * frequency
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.
422475
Angle at which light is totally reflected. \n\nThis website has a good diagram.\n\nhttp://www.physicsclassroom.com/Class/refrn/U14L3c.html
359160
what is the relationship between the relative angles of polariztion of two polarized lenses and the degree to which they obstruct light ?
642439
I assume that the light enters water through the glass since incidence is less than refracted angle also assuming refrative index of glass as 1.66 and water as 1.33\n\nWhat is the question then ? Also how can light return to where it originates ? That is dependent on the setup.
715400
Angle of line from horizontal position.