ۥ-!@ 4-5-70%%%%%% %(((((((;2)(Z)Z)Z)^)^)^)^)^)`)`)`)`)`)`)646/~)L %~)~) Refraction of Light Glass into Air PROBLEM: How is light refracted when it passes from a medium such as glass into a medium that is less optically dense, such as air? DIAGRAM: ** See Back Side of Title Page ** MATERIALS: SYMBOL 183 \f "Symbol" \s 10 \h ray box (single-slit) SYMBOL 183 \f "Symbol" \s 10 \h semi-circular glass block SYMBOL 183 \f "Symbol" \s 10 \h polar co-ordinate paper METHOD: 1. The glass block was placed on the polar co-ordinate paper, as seen in the diagram. The 0o - 180o line acted as the normal and passed through the center of the flat surface. 2. A single ray of light was directed at the curved surface of the glass, along the normal. Extra care was taken to re-assure that the ray passed through the centre of the flat surface. The angle of refraction was measured and recorded. 3. The method was repeated for angles of incidence of 10o, 20o, 30o, 40o, 50o, and 60o. Observations were then recorded in the following chart. 4. A scientific calculator was used to determine the values of the sines of the angles of incidence and refraction. 5. The ratio sin i/sinR for each pair of angles was calculated. OBSERVATIONS: Observation Angle of incidence i Angle of refraction R sin i sin R sin i sin R  1 0o 0o 0 0 -E-  2 10o 7o 0.17 0.12 1.42  3 20o 14o 0.34 0.24 1.41  4 30o 20o 0.5 0.36 1.46  5 40o 27o 0.64 0.45 1.42  6 50o 32o 0.76 0.52 1.46  7 60o 36o 0.87 0.59 1.47   ANALYSIS: Sources of Error: The sources of error encountered in this experiment where the overall shape of the polar co-ordinate paper, the overall shape of the paper used to create a single ray of light, the other interfering sources of light, and the stability of the person's hand directing the ray box. The polar co-ordinate paper could affect the results because it was not new. The condition of the of the paper was second-hand, and it had many wrinkles in it. These problems could cause the ray of light to be measured incorrectly. The paper used to create a single ray of light could affect the results because the slit in the paper was cut by hand, and had obviously been man-handled before. Therefore the slit may become too large allowing more than the required amount of light to enter the semi-circular glass block. Other interfering sources of light could affect the results in this lab by giving one the impression that the angle of incidence, refraction, or reflection are in a location in which they actually aren't. Directing the ray box manually can affect the results because the angle it is lined up on is only as good as the person's hand eye coordination. Most of these errors could be minimized if the products used are in new conditions. These errors could be minimized even further by working in a room with only one source of light, the ray box. Finally, the use of an automated ray box, which can direct itself, would help to further reduce errors. This experiment could be improved by using all new materials, and as little human involvement as possible. QUESTIONS 15.1 1. When light travels from air to glass with an angle of incidence of 0o, there is no refraction on the other end, where the light exits. 2. When light travels from air to glass at an angle of incidence greater that 0o, it bends towards the normal. 3. In relation to the normal, the incident and refracted rays are located on opposite sides. 4. The angle of refraction is always smaller than the angle of incidence in each case. 5. The sin i/sin R for all angles of incidence greater than 0o, is generally constant. 6. If light travels from air to glass, it bends away from the normal. CONCLUSION: In conclusion, light is refracted towards the normal when it passes from air into an optically denser medium, like glass. This result is valid, even with a small allowance for error, because this is how the mechanics of light work. curvedAngle of Comments 0o 10o14 20o30 30o47 40o74 50o 60o 00undno refraction0.170.240.710.340.50.680.50.730.680.640.960.69ray was broken, spectrum was visible0.770 m/9^hjk$  !BCZ[abhikorvwy{            N 46;<@AU[fy -.:G135;DLMNOPQTUXYZ[\]^_`bcdefghijlmnopqrstvwxyz{|}~    Z-/9\^j&Z!.BEZ]dkry{ĺR5lKV<: jh h h h h h!hh!h!h!!!!6 {~ "(.469>CIOUWY[ĕĕĕĕĕĕĕ!h.lKV<: jh h h h h h3[fy h rA8:G135 !W!!""n##>$r$$$$$$$V%X%%%¹°l4 ,h h!h!h!h!!!!h!h(!$2F +"޵  i5 {[5 ;Times New Roman Symbol&ArialAdjutant-Normal #C`999"h٥٥ /Andrew LikakisAndr      ' 5 Y Z !!!"""##$I$J$n$o$r$$$$$$$$$$$$$$=%T%V%X%`%g%t%}%%%   Uundno refraction0.870 undno refractionQUESTIONS 15.21. When the angle of incidence was 0o, the light went right through the glass block, and no refraction occured. 2. When light travels from glass into air, at an angle other than 0o, it bends away from the normal. 3. The angle of refraction is always greater than the angle of incidence. 4. The incident and refracted rays are locaoted on on opposite sides of the normal. 5. The value of the sin i/sinR ratio for glass-air interface is approximatley 0.68. 6. The relationship between this value and the value determined in step 5 in the previous investigation is that they are reciprocals. 7. There is no refraction at the curved surface in all cases because rays travelling along the diameter of a circular piece of glass are not refracted at either boundary, since the angles of incidence are zero at each interface. 8. Other phenomenon which occurs increasingly as the angle of incidence increases is the light spectrum. 9. Above 45o, none of the light refracts once it reaches the boundary between the glass and the air. 10. At 42.8o, the angle of refraction reaches 90o. sin i / sin 90 = 0.68 sin i = 0.68 1 i= sin-10.68 i=42.8o   In conclusion, light is refracted, when it passes from a medium such as a glass into a medium that is less optically dense, such as air, away from the normal. occurredlocatedapproximatelytraveling 44 4fXQbvy $)/3>CHNTXchntz&@  k \2!4!!(:(:(:(:(:(:(:(:!!!!!!!!!!!!!!!!!,!!!% {[% ;Times New Roman Symbol&ArialAdjutant-Normal #C`4999 Bd8    SJ4u.je  Bd8    SJ4u.je  Bd   SJ4u.je  Bd8    SJ4je  Bd8    SJ4u.je  Bd8    SJ4u.je  hL  Qbkwy  "$')-/13456789:;<>@CFHLNRTVXYZ[\]^_`acehlnrtxz  T ] n GT-6KZ !BC.0245;7,E]dkDw{LMNOPQT| U~ XYZ[\]^_ ` bcdefghi j lmnopqrs t vwxyz{|} ~    & , 2 467 : A G M S UW'  5 Z  X%   !`%!!g%!""t%""$J$}%p$r$$$$$$$8 :$$$=%%1V%"h٥w٥u٥a Y/Andrew LikakisAndrew Likakis