lab 6 physics

PHYS 2092 Lab 6: Refraction and Reflection Purpose (2 points): The purpose of the experiment is to determine how the rays of light are reflected and refracted when they go from one medium to another. Theory (2 points): We will determine the angles of reflection, refraction, and critical angle by using the law of reflection and Snell’s law. We will first use the equation  i =  r to determine the predicted angle of reflection. Using Snell’s law with the following equation we will be able to predict the angle of refraction in the two known media: n 1 sin ( θ 1 ) = n 2 sin ( θ 2 ) We will also calculate the index of refraction using the following equation: n 1 = sin θ 1 n 2sin θ 2 Finally, we will determine the critical angle using the following equation: n 1 sin θ c = n 2 sin 90 →θ c = sin − 1 ( n 2 n 1 ) Data (5 points): Data Table 6.1 Two Known Media Incident θ 1 = 25 o Predicted θ r : Experimental θ r : % difference: Predicted θ 2 : Experimental θ 2 % difference: 16.36  16.4  0.244% Unknown Medium Incident θ 1 = 35 o n 2 = 1.500 θ 2 : 67.6  n 1 : 0.353 Critical angle n 1 :1.333 n 2 : 1.500 Predicted θ c : Experimental θ c % difference: 62.70  62.8  0.159% Data Analysis (10 points) Insert relevant Data Analysis work. Two Known Media Show Work for Predicted Angle of Reflection θ i = θ r 25 o = 25 o Show Work for Predicted Angle of Refraction n 1 sin ( θ 1 ) = n 2 sin ( θ 2 ) 1.000sin (25  ) = 1.500 sin (θ 2 ) sin ( θ 2 ) = ( 1.000 1.500 ) x sin ( 25 ° ) = ¿¿ 0.2817 θ 2 = arcsin ( 0.2817 ) = 16.36 °

Percent Difference %Difference = | 1 st value − 2 nd value | average of the two × 100% %Difference = 16.4 − 16.36 16.38 x 100% = 0.244 % Unknown Media List Given and Unknown n 1 = Unknown n 2 = 1.500 Incident angle = 35  Refracted angle measured = 67.6  Show Work for Index of Refraction n 1 = sin θ 1 n 2sin θ 2 = sin 35 ° ( 1.5 ) sin 67.6 = 0.353 Critical Angle List Given and Unknown n 2 = 1.500 n 1 = 1.333 θ c = unknown Show Work for Critical Angle n 1 sin θ c = n 2 sin 90 →θ c = sin − 1 ( n 2 n 1 ) θ c = sin − 1 ( 1.333 1.500 ) = 62.70  %Difference = | 1 st value − 2 nd value | average of the two × 100% %Difference = 62.8 − 62.70 62.75 x 100% = 0.159% Summary of Results (2 points Insert relevant screenshots. Two Known Media Incident θ 1 = 25 o *** Default was air at 1.000 not glass as stated in instructions ***

Discussion (3 points) From your observations, does light refract towards the normal or away from the normal when it travels from one medium into another medium with a higher index of refraction? When the light enters a medium with a higher refraction index like the first experiment from air to glass the light refracts towards the normal line. If the light enters a medium with a lower refraction index it will reflect away from the normal line. Conclusion (3 points) In conclusion we can see how light entering a medium depends on the refraction index to determine the reflection. In the experiment my percent differences were minimal indicating a low precent of error in using the equations VS experiments. Overall Quality of Report (3 points) All part of the lab report are included and correctly organized, section headings are properly labeled, entire report is clear and legible, with correct grammar and spelling.