Please provide deep explaining of concepts, theories, and formulas that apply to

Question

Please provide deep explaining of concepts, theories, and formulas that apply to each problem. I have a physics final coming up and I need to be prepared, these are the practice questions. thank you.

Please draw sketches that help visualize problem solving as well, thank you.

y Notoo Ask Your Teachee A ray of light is incident on the surface ofe block of dear ice 8t·angle of 54 with the norr el Part of the light is enected end pertis efracted find the engle bet een the reflected end refracted light Need Help? PdN po ies Th" light beam shown 'n tie ure bolaw mabec an angle o # = 25.4 with the normal irie NW. In th. Iwned cll. Drte min" th" a glee fi an 0' The. refractiv-Index ar linseed il 16 1 48. The index of refraction for red ligint in water is 1.331 and that for blue light is 1.340. I a ray of white light anters the water at an angle of incidenes of 67.70, what are the underwater anglea of refraction for the blue and red compenenita of the lig? (Enter your anawers to at leaat two decimal places. (a) blue component (a) blue como (b) red component Need Help? eth

Explanation / Answer

1. let the angle of incidence be i = 54 degree
   then from laews of reflection
   angle of reflection = r = i
   and angle of refraction be r'
   from snell's law, sin(i)/sin(r') = n ( where n is refractive index of clear ice)

   also, angle between relected and refracted ray = 90 - r + 90 - r' = 180 -(r + r')
   so, n = 1.309
   so, r = i = 54 deg
   and sin(54)/sin(r') = 1.309
   r' = 38.17 degree

   so angle between reflected and refractedf rays = 180 - (54+38.17) = 87.83 degrees

2. given refractive index of lineseed oil, n = 1.48
   then from snell's law
   sin(theta)/sin(phi) = n
   but phi = 26.4 deg
   so sin(theta) = 1.48*sin(26.4)
   theta = 41.15 degree

   siniliarly
   sin(phi)/sin(theta') = 1.33/1.48
   sin(theta') = 1.48*sin(26.4)/1.33
   theta' = 29.655 deg

3.   n-red (water) = 1.331
   n-blue (water) = 1.340
   angle of incidence, i = 67.30 deg
   a) Angle of refraction for blue component = r
   from snell's law
   sin(67.30)/sin(r) = 1.340
   r = 43.508 degree
   a) Angle of refraction for red component = r'
   from snell's law
   sin(67.30)/sin(r') = 1.331
   r = 43.877 degree
   

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