Data & Analysis Procedure A Incident Angle 1 Refracted Angle 1 Incident Angle 2 Refracted Angle...

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

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

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

Solutions

Expert Solution

genius_generous answered 1 month ago

Next > < Previous

Related Solutions

Using detailed incident angle and refracted angle measurement data. Determine the refractive index of each material...

Using detailed incident angle and refracted angle measurement data. Determine the refractive index of each material and compare with the theoretical values. Data group 1 Theoretical index: 1.33 incident angle: 10, 20, 30, 40, 50, 60, 70, 80 refracted angle: 7.2, 14.63, 22.03, 28.92, 35.9, 40.67, 44.12, 48.72 Data groups 2 Theoretical index: 1.45 incident angle: 10, 20, 30, 40, 50, 60, 70, 80 refracted angle: 6.86, 13.74, 19.65, 25.72, 32.5, 35.59, 42.16, 42.65 Data group 3 Theoretical index: 1.22...

1. How does the brightness of the refracted ray change when the incident angle (from glass...

1. How does the brightness of the refracted ray change when the incident angle (from glass to air) changes from less than critical angle to greater than critical angle? a. brighter b. dimmer c. unchanged d. dimmer and then disappear 2. Trying to picture an electroscope you made in the lab, on rubbing an rubber rod with fur, the electrons from the fur get transferred to the rubber rod. While holding the rubber rod near the electroscope (not touching the...

What determines how light bends? Use the words: index refraction, incident angle, reflected angle, refracted angle,...

What determines how light bends? Use the words: index refraction, incident angle, reflected angle, refracted angle, and color.

Light moves from inside an ice cube to air. At what angle was the light incident upon the ice/air boundary if the refracted angle was 35°?

Question 20 Light moves from inside an ice cube to air. At what angle was the light incident upon the ice/air boundary if the refracted angle was 35°? The index of refraction for the air is 1.0 and 1.4 for the ice.Question 20 options:53°35°12°66°24°Question 203An object is placed 10 cm in front of a concave mirror of focal length 5 cm. The image produced isQuestion 23 options:Virtual and uprightNone of the aboveReal and invertedReal and uprightVirtual and inverted

When reaching a boundary between two media, an incident ray is partially reflected and partially refracted....

When reaching a boundary between two media, an incident ray is partially reflected and partially refracted. The ray is travelling from media 1 to media 2. The indexes of refraction for the two media are n1 and n2, respectively. At what angle of incidence is the reflected ray perpendicular to the refracted ray? Express your answer in terms of the variables n1 and n2.

Rotate the stage slightly such that the incident angle is just below the critical angle. Which...

Rotate the stage slightly such that the incident angle is just below the critical angle. Which color appears first. Why is this? How does the index of refraction depend on the wavelength of light?

A beam of white light is incident on the surface of a diamond at an angle...

A beam of white light is incident on the surface of a diamond at an angle ?a.(Figure 1) Since the index of refraction depends on the light's wavelength, the different colors that comprise white light will spread out as they pass through the diamond. The indices of refraction in diamond are nred=2.410 for red light and nblue=2.450 for blue light. The surrounding air has nair=1.000. Note that the angles in the figure are not to scale. Part A Calculate vred,...

A beam of light in air is incident at an angle of 24.0° to the surface...

A beam of light in air is incident at an angle of 24.0° to the surface of a rectangular block of clear plastic ( n = 1.46). The light beam first passes through the block and re-emerges from the opposite side into air at what angle to the normal to that surface?

Light in a vacuum is incident on a transparent glass slab. The angle of incidence is...

Light in a vacuum is incident on a transparent glass slab. The angle of incidence is 37.0

A light ray of wavelength 589 nm is incident at an angle θ on the top...

A light ray of wavelength 589 nm is incident at an angle θ on the top surface of a block of crown glass surrounded by air, as shown in the figure below. (The index of refraction of crown glass is 1.52.) (a) Find the maximum value of θ for which the refracted ray will undergo total internal reflection at the left vertical face of the block. ° (b) Repeat the calculation for the case in which the crown glass block...

Subjects