Data & Analysis

Procedure A

Incident Angle 1	Refracted Angle 1	Incident Angle 2	Refracted Angle 2	Incident Angle 3	Refracted Angle 3
90-68 = 22 degrees	90-74 = 16 degrees	90-55 = 35 degrees	90-60 = 30 degrees	90-44 = 46 degrees	90-50 = 40 degrees

Procedure B – concave

Incident Angle 1	Refracted Angle 1	Incident Angle 2	Refracted Angle 2	Incident Angle 3	Refracted Angle 3
99-90 = 9 degrees	101-90 = 11 degrees	90-90 = 0 degrees	90-90 = 0 degrees	90-80 = 10 degrees	98-90 = 8 degrees

Procedure B – convex

Incident Angle 1	Refracted Angle 1	Incident Angle 2	Refracted Angle 2	Incident Angle 3	Refracted Angle 3
96-90 = 6 degrees	102-90 = 12 degrees	90-90 = 0 degrees	90-90 = 0 degrees	96-90 = 6 degrees	100-90 = 10 degrees

Procedure C

Incident Angle 1	Refracted Angle 1	Incident Angle 2	Refracted Angle 2
90-60 = 30 degrees	90-75 = 15 degrees	90-75 = 15 degrees	115-90 = 25 degrees

Procedure D

Focal Point 1 (converging)	Focal Point 2 (diverging)
5 cm	-4.7cm

Procedure E

The angle of Red is smaller than the angle of Blue.

Procedure F – critical angle

Incident Angle 1

90-51 = 39 degrees

Questions/Applications

1. In procedure A & B, compare the incidentand reflected angles. Find a % error.
2. For the procedure, C find the average index ofrefraction for the plastic block. (“Average” because you havetwo sets of the incident and refracted angles that you can applySnell’s law too.)
3. For procedure D which lens produced a realimage and which produced a virtual image and why?
4. For procedure E what color of the light wasdispersed the most (dispersed meaning the color-dependent angle ofrefraction). Is your result consistent with equation (n = speed oflight/frequency of light x wavelength? Explain your answer.
5. Use the critical angle in procedure F to findthe index of refraction of the plastic half-circle. Compare this tothe index of refraction found in question 2. Find a %difference.

HERE IS THE DATA FOR THE QUESTIONS. PLEASE ANSWER THEMFOR ME! THANKS!