Pre-Lab QuestionsUse Figure to answer Pre-Lab Questions and Figure Light indicated by the arrow traveling 4779- CoachOnCouch

PHYS 253 Physics Lab for Engineers

Question.4779 - Pre-Lab QuestionsUse Figure 7 to answer Pre-Lab Questions 1 and 2.Figure 7: Light (indicated by the arrow) traveling through three materials, A, B, and C.Rank the index of refraction in each of the materials from highest to lowest. If two quantities are equal, denote they are equal with an equal sign.In terms of determining the ranking of refractive indices, observing the pattern of bending of light, in scenario the light bends towards the normal, it is entering a material with a higher index of refraction. If it bends away from the normal, it is entering a material with a lower index; so here the light seems to be normal at material C and A.Highest ____B__ ___A____ __C___ Lowest Rank the speed of light in each of the materials from fastest to slowest. If two quantities are equal, denote they are equal with an equal sign. Speed of light is inversely proportional to the index of refractionv= CnFastest ____C__ ___A_______B____ SlowestWater has an index of refraction of 1.33. What is the angle of refraction if a laser beam strikes the surface of the water at 30 degrees above the surface of the water? Show your work.Using Snellâs law n1sinÎ¸1= n2sinÎ¸21.00*sin60o= 1.33*sinâ¡(Î¸2)0.8660=1.33*sinÎ¸2â sinÎ¸2 = 0.86601.33=0.6519Î¸2=sin-1(0.6519)â40.8oThe critical angle for a diamond surrounded by air is 24.4 degrees. What is the index of refraction for diamond? Show your work.SinÎ¸C=n2n1, n2=1.00 airSin24.4o=1.00n1 ân1=1.00Sin24.4o=1.000.4130 â2.42A diver is testing out a new waterproof camera on a scuba diving adventure. If the diver is 10 m.Firstly, considering pressure at 10 meters depth (contextual but common)P= P0+pghTaking P0=101325Pa atmospheric pressure, p=1000 kgm3, g=9.81 ms2, h=10mP=101325+10009.8110=101325+98100=199425 PaTotal pressure at 10 m depth â 199,425 Pa â 1.97 atmConsidering apparent depth of an object due to refraction, when viewing from water into air n = 1.33 to 1.00 the object appears closer to the surface, now utilizing the formula for apparent depth: Apparent depth= Real Depthn=10 m1.33=7.52 m, So the object will appear to be at 7.52 meters above the diver rather than the actual 10 meters.In conclusion, pressure at 10 m depth: ~1.97 atm, apparent position of object ~ 7.52, above diver due to refraction from water to air. Experiment 1: law of reflectionData SheetTable 2: Laser and Mirror Reflection DataIncident Angle, Î¸i (Degrees)Reflected Angle, Î¸T (Degrees)005510101515202025253030353540404545505055556060656570707575Post-Lab QuestionsDescribe the relationship between the incident and reflected angle.The angle of incidence equals the angle of reflection for all anglesÎ¸i=Î¸rIs this true for all materials? What happens when light is reflected off of pencil lead on a piece of paper?Yes, the law of reflection holds on all surfaces, however, diffuse reflection occurs on rough surfaces like pencil lead on paper, scattering the light in many directions.Experiment 2: snellâs lawData SheetTable 3: Laser and Refraction Cell DataIncident Angle, Î¸i (Degrees)Reflected Angle, Î¸T (Degrees)SinÎ¸i (Degrees)SinÎ¸T (Degrees)000.00000.000053.80.08720.0662107.50.17360.13051511.20.25880.19452014.90.34200.25712518.50.42260.31733022.10.50000.37503525.50.57360.43134028.90.64280.48484532.10.70710.5314Post-Lab QuestionsWhat is the critical angle for a laser beam travel from air to water?Not applicable since no critical angle when entering a denser medium (water); total internal reflection only occurs from higher to lower refractive index.There are two boundaries where the laser beam changes medium. Describe the two boundaries and determine which one causes the light to refract. Air to water: Refraction occurs (light bends toward normal).Water to air: Refraction or total internal reflection occurs depending on angle.The two mediums in this experiment are water and air. Which medium is the medium for the incident ray? Which medium is the medium for the refracted ray?Incident ray: AirRefracted ray: WaterThe index of refraction for water is n = 1.33. Use this value to calculate the theoretical critical angle for the laser travelling from water to air. Compare this value to the one you experimentally determined with a percent error calculation. Show your work.Sin(Î¸C)=n2n1=48.75-4748.75*100=3.59%Complete Table 3 by computing the sine of the incident and refracted angle.Plot a graph of sin(Î¸i) versus sin(Î¸r). Construct your graph on a computer program such as Microsoft Excel ®.Draw in a line of best fit on your graph.Hint: You want the y-intercept to be zero.Estimate the slope of the line and compare the slope to the index of refraction for water (n = 1.33) with a percent error calculation. Show your work and explain any discrepancies between the experimental index of refraction and the actual index of refraction.From earlier calculations (Experiment 2), we have the following values for sin(Î¸i) and sin(Î¸r)Incident Angle, Î¸i (Degrees)Reflected Angle, Î¸T (Degrees)SinÎ¸i (Degrees)SinÎ¸T (Degrees)000.00000.000053.80.08720.0662107.50.17360.13051511.20.25880.19452014.90.34200.25712518.50.42260.31733022.10.50000.37503525.50.57360.43134028.90.64280.48484532.10.70710.5314Considering graph of sin(Î¸i) (y-axis) vs. sin(Î¸r) (x-axis), the slope should equal the index of refraction ð2/ð1 = nwater/nair = 1.33Choosing two well separated points Point A: X = Sin(Î¸r) = 0.1305, Y = Sin(Î¸i) = 0.1736Point B: X = 0.5314Y = 0.7071Considering slope = ÎYÎX=(0.7071-0.1736)(0.5314-0.1305)=0.53350.4009=1.33In terms of percent error calculationPercent error = (Experimental-Actual)Actual*100=1.33-1.331.33*100=0% In terms of discrepancies (if slope â 1.33)In scenario the slope is 1.29Percent error = (Experimental-Actual)Actual*100=1.29-1.331.33*100=3.01% However, the possible reasons for discrepancy (based on my understanding): possibly due to human error in reading angles, imperfect alignment of laser beam and the protractor, beam width causing uncertainty in exact point of incidence/refraction; lastly smaller surface ripples in water and inaccurate sin () values if rounded too early. Do you always observe a reflected ray of light as you increase the incident angle? What happens to the intensity of the reflected light as you increase the incident angle and eventually cross the critical angle?Based on the experiment, yes â a reflected ray is always observed at the boundary between two media, regardless of incident angle which is due to partial reflection that occurs in every case of light encountering a boundary. As the incident angle increases, the intensity of the reflected light increases because the angle between the incident ray and the normal increases, and the fraction of light reflected increases per Fresnel equations. When the incident angle exceeds the critical angle 48.75o for water-to-air total internal reflection occurs where no refraction happens, all light is reflected back into the medium and reflected light becomes fully intense and undiminished; such principle behind fiber optics and underwater periscopes light is confined by total internal reflection.

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x In xxxxx of xxxxxxx error xxxxxxxxxxxxxxxxxx error xxxxxxxxxxxxxxxxxxx Actual x In xxxxx of xxxxxxxxxxxxx if xxxxx In xxxxxxxx the xxxxx is xxxxxxx error xxxxxxxxxxxxxxxxxxx Actual x However xxx possible xxxxxxx for xxxxxxxxxxx based xx my xxxxxxxxxxxxx possibly xxx to xxxxx error xx reading xxxxxx imperfect xxxxxxxxx of xxxxx beam xxx the xxxxxxxxxx beam xxxxx causing xxxxxxxxxxx in xxxxx point xx incidence xxxxxxxxxx lastly xxxxxxx surface xxxxxxx in xxxxx and xxxxxxxxxx sin xxxxxx if xxxxxxx too xxxxx Do xxx always xxxxxxx a xxxxxxxxx ray xx light xx you xxxxxxxx the xxxxxxxx angle xxxx happens xx the xxxxxxxxx of xxx reflected xxxxx as xxx increase xxx incident xxxxx and xxxxxxxxxx cross xxx critical xxxxx Based xx the xxxxxxxxxx yes x reflected xxx is xxxxxx observed xx the xxxxxxxx between xxx media xxxxxxxxxx of xxxxxxxx angle xxxxx is xxx to xxxxxxx reflection xxxx occurs xx every xxxx of xxxxx encountering x boundary xx the xxxxxxxx angle xxxxxxxxx the xxxxxxxxx of xxx reflected xxxxx increases xxxxxxx the xxxxx between xxx incident xxx and xxx normal xxxxxxxxx and xxx fraction xx light xxxxxxxxx increases xxx Fresnel xxxxxxxxx When xxx incident xxxxx exceeds xxx critical xxxxx o xxx water-to-air xxxxx internal xxxxxxxxxx occurs xxxxx no xxxxxxxxxx happens xxx light xx reflected xxxx into xxx medium xxx reflected xxxxx becomes xxxxx intense xxx undiminished xxxx principle xxxxxx fiber xxxxxx and xxxxxxxxxx periscopes xxxxx is xxxxxxxx by xxxxx internal xxxxxxxxxx