## Tuesday, January 14, 2020

### The Refractive Index

The refractive index Aim: The aim of this experiment is to find the refractive index of a glass prism. In this experiment, the independent variable is the angle of incidence, and the dependent variable is the angle of refraction. Theory: SnellÃ¢â‚¬â„¢s law relates the angle of incidence and refraction to the ratio of the velocity of the wave in the different media. The formula for SnellÃ¢â‚¬â„¢s law is the following: Sin isinr = v1v2 = n Where i is the angle of incidence, r is the angle of refraction and v1 and v2 are the velocities of the wave in different media and n is the refractive index.Light refracts when it passes from one medium to another. The ratio of the velocity of light in the two media is called the refractive index. Materials and method: For this experiment we used a half glass circle attached on the center of a laminated paper with a drawn circle around it, a blue/violet laser with a wavelength 447nm and a wood block. First we started by placing the flat side of the half glass circle attached to the paper in front of the laser. Depending on the angle we wanted to find, 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 10Ã‚ °, 20Ã‚ °, 30Ã‚ °, 40Ã‚ °, 50Ã‚ ° and 60Ã‚ °. First we measured the angle of incidence, where we placed the wood block perpendicular to the ray. To control the variables, the laser should have the same wavelength for all the angles to get the same refractive index and the ray should hit the center of the glass circle, so to check that the ray hits the center of the glass circle, we placed a wood block at the angle of reflection to see if the angle of reflection is the same as the angle of incidence, because we know that the angle of incidence is equal to the angle of reflection.Another thing which makes 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 ce nter 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 close where the ray hits the wood block, to see more precise where the angle of refraction is.We repeated this method for all the different angles of incidence and repeated every angle two times. D 2 1 2 You write Ã¢â‚¬Å"to control the variablesÃ¢â‚¬  which variables? You should mention the wavelength and the hitting the center explicitly as variables to be controlled and why. Results: Angle of incidence Ã‚ ± 0. 1Ã‚ °| Angle of refraction1 Ã‚ ± 0. 1Ã‚ °| Angle of refraction2 Ã‚ ± 0. 1Ã‚ °| Angle of refraction3 Ã‚ ± 0. 1Ã‚ °| 10Ã‚ °| 6. 9Ã‚ °| 7. 1Ã‚ °| 7. 0Ã‚ °| 20Ã‚ °| 13. 6Ã‚ °| 13. 5Ã‚ °| 13. 5Ã‚ °| 30Ã‚ °| 20. 0Ã‚ °| 20. 1Ã‚ °| 20. 0Ã‚ °| 40Ã‚ °| 25. 6Ã‚ °| 25. 8Ã‚ °| 25. 7Ã‚ °| 50Ã‚ °| 30 . 7Ã‚ °| 30. Ã‚ °| 30. 8Ã‚ °| 60Ã‚ °| 35. 9Ã‚ °| 35. 9Ã‚ °| 36. 0Ã‚ °| Example: First we find the average and uncertainty for the angle of refraction: 7. 2- 6. 8 2 = Ã‚ ± 0. 2Ã‚ ° Angle of incidence Ã‚ ± 0. 1Ã‚ °| Average angle of refractionÃ‚ ± 0. 2Ã‚ °| 10Ã‚ °| 7. 0Ã‚ ° | 20Ã‚ °| 13. 5Ã‚ ° | 30Ã‚ °| 20. 0Ã‚ ° | 40Ã‚ °| 25. 7Ã‚ ° | 50Ã‚ °| 30. 8Ã‚ ° | 60Ã‚ °| 35. 9Ã‚ ° | 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) Ã¢â‚¬â€œ sin (9. )sin(7. 2) )/2 =0. 045 ? Ã‚ ± 0. 05 sin(10Ã‚ °) sin(7. 0Ã‚ °) = 1. 42 Ã‚ ± 0. 05 Angle of incidence Ã‚ ± 0. 1Ã‚ °| Angle of refraction Ã‚ ± 0. 2Ã‚ °| Refractive index| 10Ã‚ ° | 7. 0Ã‚ °| 1. 42 Ã‚ ± 0. 05| 20Ã‚ °| 13. 5Ã‚ ° | 1. 47 Ã‚ ± 0. 03| 30Ã‚ °| 20. 0Ã‚ ° | 1. 46 Ã‚ ± 0. 02| 40Ã‚ °| 25. 7Ã‚ ° | 1. 48 Ã‚ ± 0. 01| 50Ã‚ °| 30. 8Ã‚ ° | 1. 50Ã‚ ± 0. 01| 60Ã‚ °| 3 5. 9Ã‚ ° | 1. 48 Ã‚ ± 0. 01| Refractive index Intervals: Angle of incidence Ã‚ ± 0. 1Ã‚ °| Refractive index intervals| 10Ã‚ ° | 1. 37 Ã¢â‚¬â€œ 1. 47| 20Ã‚ °| 1. 44 Ã¢â‚¬â€œ 1. 50| 30Ã‚ °| 1. 44 Ã¢â‚¬â€œ 1. 48| 40Ã‚ °| 1. 47 Ã¢â‚¬â€œ 1. 49| 50Ã‚ °| 1. 49 Ã¢â‚¬â€œ 1. 51| 60Ã‚ °| 1. 47 Ã¢â‚¬â€œ 1. 49| DCP 2 2 2 Conclusion:From the table we can see that there is no interval, where at least one number from each interval is included. The consequences of the small angles are more serious than the bigger angles. SnellÃ¢â‚¬â„¢s law states that no matter 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 10Ã‚ °), I can state that SnellÃ¢â‚¬â„¢s 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 also be some misreading when reading the small angles, that has leaded to that the small angles of incidenceÃ¢â‚¬â„¢s results are a bit 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 thing to improve the method is by using a Vernier 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