The Refractive Index

The deflective major power Aim The aim of this experiment is to key the refractive superpower of a field fruitcake prism. In this experiment, the independent variable is the tap of relative relative incidence, and the dependent variable is the travel of deflexion. surmisal Snells law relates the burden of incidence and refraction to the ratio of the velocity of the roll up in the different media. The formula for Snells law is the following Sin isinr = v1v2 = n Where i is the topple of incidence, r is the angle of refraction and v1 and v2 atomic number 18 the velocities of the wave in different media and n is the refractive ability.Light refracts when it passes from unmatchable medium to an new(prenominal). The ratio of the velocity of strike in the 2 media is c totallyed the refractive index. Materials and regularity For this experiment we used a hotshot-half codswallop rope attached on the tenderness of a laminated paper with a displace circle around it, a blue/violet optical maser with a wavelength 447nm and a timber block. First we started by placing the flat side of the half scrap circle attached to the paper in front of the laser. Depending on the angle we treasu reddish to arrest, we used the drawn circle on the paper to decide where to put the laser on the half side of the drawn circle.The angles of incidence we used were 10, 20, 30, 40, 50 and 60. First we heedful the angle of incidence, where we pose the wood block orthogonal to the ray. To project the variables, the laser should have the same(p) wavelength for all the angles to get the same refractive index and the ray should hit the promenade of the glass circle, so to check that the ray hits the center of the glass circle, we hardened a wood block at the angle of observation to go steady if the angle of reflection is the same as the angle of incidence, because we accredit that the angle of incidence is equal to the angle of reflection.Another function which m akes it easier to hit the center of the glass circle is by placing a paper on the flat side of the circle and see if the ray hits the center of the circle and by placing the wood block perpendicular. Then we measured the angle of the refraction on the other half of the drawn circle, where we again placed the wood block perpendicular. We measured the angle of refraction by looking perpendicular down from the wood block, to see constrictive where the ray hits the wood block, to see more than(prenominal) precise where the angle of refraction is.We repeated this order for all the different angles of incidence and repeated every angle two times. D 2 1 2 You write to control the variables which variables? You should mention the wavelength and the hit the center explicitly as variables to be controlled and why. Results pitch of incidence 0. 1 move of refraction1 0. 1 Angle of refraction2 0. 1 Angle of refraction3 0. 1 10 6. 9 7. 1 7. 0 20 13. 6 13. 5 13. 5 30 20. 0 20. 1 20. 0 4 0 25. 6 25. 8 25. 7 50 30. 7 30. 30. 8 60 35. 9 35. 9 36. 0 Example First we find the amount and uncertainty for the angle of refraction 7. 2- 6. 8 2 = 0. 2 Angle of incidence 0. 1 Average angle of refraction 0. 2 10 7. 0 20 13. 5 30 20. 0 40 25. 7 50 30. 8 60 35. 9 The refractive index We know that the formula is sinisinr = v1v2 = refractive index, so by applying the information we know to the formula, we can find the refractive index. Example Uncertainty for refractive index ( sin(10. 1)sin(6. 8) sin (9. )sin(7. 2) )/2 =0. 045 ? 0. 05 sin(10) sin(7. 0) = 1. 42 0. 05 Angle of incidence 0. 1 Angle of refraction 0. 2 refractive index 10 7. 0 1. 42 0. 05 20 13. 5 1. 47 0. 03 30 20. 0 1. 46 0. 02 40 25. 7 1. 48 0. 01 50 30. 8 1. 50 0. 01 60 35. 9 1. 48 0. 01 deflective index Intervals Angle of incidence 0. 1 Refractive index intervals 10 1. 37 1. 47 20 1. 44 1. 50 30 1. 44 1. 48 40 1. 47 1. 49 50 1. 49 1. 51 60 1. 47 1. 49 DCP 2 2 2 codaFrom the tab le we can see that in that location is no interval, where at least one number from each interval is included. The consequences of the modest angles are more serious than the large angles. Snells law put ins that no bet what the angle of incidence is, the refractive index would be the same. From the results I gained (disregarding the angle of incidence equals to 10), I can state that Snells law is confirmed in this case. Evaluation The method has some weaknesses. The glass prism is not exactly in the center of the drawn circle, which is why the results are not quite correct.There might likewise be some mis discipline when reading the microscopical angles, that has leaded to that the small angles of incidences results are a arcsecond uncommon and almost an outliers, but overall reading the angles could be one of the errors too. Suggestions It would be better to glue the glass prism more precise in the center of the circle, so that the result would be more precise. Another thi ng to improve the method is by using a vernier scale gauge to measure the size of small distances more accurate. CE 1 2 2 The one because the meaning of the red sentence is not clear very well, this is your best up to now. grade 7