Week Eleven Lab Portal

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Apr 3, 2024

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Renad Nofal PHYS - 2320 Prof. Gucik Nov 21 2023 Week Eleven Lab Portal Model No. 0S-8515C Experiment 4: Snell’s Law Experiment 4: Snell’s Law Required Equipment from Basic Optics System Light Source Trapezoid from Ray Optics Kit Other Required Equipment Protractor White paper Purpose The purpose of this experiment is to determine the index of refraction of the acrylic trapezoid. For rays entering the trapezoid, you will measure the angles of incidence and refraction and use Snell’s Law to calculate the index of refraction. Incident ray ! :4,_____-Normal to surface n Surface Theory ny For light crossing the boundary between two transparent materials, Snell’s Law states Refracted ray (n;>n,) nsin 0, = n,sin 6, where 6, is the angle of incidence, 0, is the angle of refraction, and n, and n, are the respective indices of refraction of the materials (see Figure 4.1). Figure 4.1 Procedure 1. Place the light source in ray-box mode on a sheet of . white paper. Turn the wheel to select a single ray. 0 f i - 2. Place the trapezoid on the paper and position it so B the ray passes through the parallel sides as shown in Incident ray Figure 4.2. 3. Mark the position of the parallel surfaces of the Figure 4.2 trapezoid and trace the incident and transmitted rays. Indicate the incoming and the outgoing rays with arrows in the appropriate directions. Carefully mark where the rays enter and leave the trapezoid. 4. Remove the trapezoid and draw a line on the paper connecting the points where the rays entered and left the trapezoid. This line represents the ray inside the trap- ezoid. 5. Choose either the point where the ray enters the trapezoid or the point where the ray leaves the trapezoid. At this point, draw the normal to the surface. 6. Measure the angle of incidence (0;) and the angle of refraction with a protractor. Both of these angles should be measured from the normal. Record the angles in the first row of Table 4.1. 17:%{ef0 © 15

Basic Optics System Experiment 4: Snell’s Law 7. On anew sheet of paper, repeat steps 2—6 with a different angle of incidence. Repeat these steps again with a third angle of incidence. The first two columns of Table 4.1 should now be filled. Table 4.1: Data and Results Analysis 1. For each row of Table 4.1, use Snell’s Law to calculate the index of refraction, Angle of Incidence Angle of Refraction Calculated index of refraction of acrylic 40 36 1.0935 56 31 1.2480 63 54 0.9913 Average: 1.1109 assuming the index of refraction of air is 1.0. 2. Average the three values of the index of refraction. Compare the average to the accepted value (n = 1.5) by calculating the percent difference. pifference = 2.71% Question What is the angle of the ray that leaves the trapezoid relative to the ray that enters it? refraction angle 16 I1J7:C1efe ©

Model No. OS-8515C Experiment 5: Total Internal Reflection Experiment 5: Total Internal Reflection Required Equipment from Basic Optics System Light Source Trapezoid from Ray Optics Kit Other Required Equipment Protractor White paper Purpose In this experiment, you will determine the critical angle at which total internal reflec- tion occurs in the acrylic trapezoid and confirm your result using Snell’s Law. Theory | Incident ray | Reflected ray For light crossing the boundary between two transpar- 0,7 ent materials, Snell’s Law states | | nlsi.n 61 =nzsin 62 nl | Surface ) where 0, is the angle of incidence, 0, is the angle of | refraction, and », and n, are the respective indices of | | refraction of the materials (see Figure 5.1). Refracted ray (ny>n,) In this experiment, you will study a ray as it passes out | of the trapezoid, from acrylic (n = 1.5) to air (n,; = 1). If the incident angle (0,) is greater than the critical angle (6,), there is no refracted ray and total internal reflection occurs. If 6, = 6, the angle of the refracted ray (0,) is 90°, as in Figure 5.2. In this case, Snell’s Law states: Incident ray | | n sin O, = 1 sin 90° 0.~ Reflected ray | | | | | | Solving for the sine of critical angle gives: - C Refracted ray S = [ = = s1n9c = I2:X{ede © 17

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Procedure 1. Place the light source in ray-box mode on a sheet of white paper. Turn the wheel to select a single ray. Position the trapezoid as shown in Figure 5.3, with the ray entering the trapezoid at least 2 cm from the tip. Rotate the trapezoid until the emerging ray just barely disappears. Just as it disappears, the ray separates into colors. The trapezoid is correctly posi- tioned if the red has just disappeared. Mark the surfaces of the trapezoid. Mark exactly the point on the surface where the ray is internally reflected. Also mark the entrance point of the incident ray and the exit point of the reflected ray. Remove the trapezoid and draw the rays that are incident upon and reflected from the inside surface of the trapezoid. See Figure 5.4. Measure the angle between these rays using a protractor. (Extend these rays to make the protractor easier to use.) Note that this angle is twice the critical angle because the angle of incidence equals the angle of reflection. Record the critical angle here: 0 = c 87 _ (experimental) Calculate the critical angle using Snell’s Law and the given index of refraction for Acrylic (n = 1.5). Record the theoretical value here: 0.= 40.2 (theoretical) C Reflected ray Incident ray Eefracted ay Figure 5.3 Exit point— A *“Reflection Entrance point point Figure 5.4 refraction angle: 90 Calculate the percent difference between the measured and theoretical values: % difference = 1.67 Questions 1. How does the brightness of the internally reflected ray change when the incident angle changes from less than 0 to greater than 0 _? when the angle of incidence is greater than tife crifical angle the brifil"\tness increase, and the brightness of the light decreases when the angle of incidence is zero. Is the critical angle greater for red light or violet light? What does this tell you about the index of refraction? When the angle of incidence increases, the brightness of the reflected ray decreases. in the red light the critical angle is greater, because the index of refraction is greater in violet ( the higher refractive index, the smaller the critical angle) 18 I12J:%]efe ©

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