Tyson Wu_s Lab 14 Snell’s Law and the Lensmaker’s Equation

Phys 152L/272L – Physics II Laboratory Snell’s Law and the Lensmaker’s Equation Due date: Friday 11/20/20 (submission on Laulima as pdf) /100 points Instructions: Please adhere to the guidelines in the syllabus to prepare your solution. You must show all calculations clearly. If you don't, you will not receive credit. Objective [4 pts]: 1. What are the objectives of this experiment? The objective of this lab is to understand how light is refracted when it passes through a lens by applying Snell’s law and the lensmaker equation. In addition, the objective of this experiment is to also test geometric optic laws to determine how lenses work Theory [8 pts]: 2. [1 pt] Look at the image on the right. Which set of angles are the correct ones to use in Snell’s Law? a. θ 1 b. θ 2 3. [3 pts] Explain why the index of refraction of a material must be greater than 1. The index of refraction (n) of a material must be greater than 1 because the speed of light in a material (v) cannot be greater than the speed of light in air. Therefore when we apply it to the equation for the index of refraction n = c/v, we see that because v is always less than c when in a material, the n value must always be greater than 1. 4. [2 pts] The refraction of a light ray due to a lens occurs a. continuously inside the lens. b. twice: once as the ray enters the lens, and once as the ray exits the lens. c. once, along the center of the lens d. once, as the ray exits the lens

5. [2 pts] Spherical aberration occurs when a. light rays at different distances from the lens’s axis of symmetry don’t focus together. b. the lens fails to produce an inverted image. c. not all the light that enters the lens can exit the lens. d. light rays of different wavelengths don’t focus together. Procedure [12 pts]: 6. [4 pts] Explain how the “small” and “ring” apertures affect how the light reaches the lens. The small aperture only allows light from the center to go through to the lens, whereas the ring aperture only allows light from the outside into the lens. 7. [4 pts] How does the use of apertures allow us to measure the spherical aberration? Hint: How will the focal length f measured with one aperture compare to the focal length f measured with the other aperture? Assume no spherical aberration. Does your answer change for a lens with spherical aberration? To determine the correct image distance s’ we must move the screen until we produce the most focused image. We have found the correct image distance when the image is at its most focused position, by adjusting the aperture 8. [4 pts] How do we know when we’ve found the correct image distance, s ′ ? What piece of the apparatus do we move to find s ′ ? To determine the correct image distance s’ we must move the screen until we produce the most focused image. We have found the correct image distance when the image is at its most focused position, by adjusting the aperture Data Analysis [38 pts]:

10. [11 pts] Plot vs. from the ring aperture data. Calculate the focal length of the outer s ′ s rays, f , and its error, . ring δ f ring - m (slope) = -0.9399 +/- 0.00564 - B (y-intercept) = 0.04501 +/- 0.000125 - ( ) f ring = 2 1 1 B − B m - ( ) f ring = 2 1 1 0.04501 − 0.04501 −0.9399 - 1.54965 cm f ring = 2 - f δ ring = 1 2 B 2 √ (1 ) δ B δ m − m 2 2 + B 2 2 - f δ ring = 1 2(0.04501) 2 √ (1 − .9399)) (0.000125) 0.04501) (0.00564) − ( 0 2 2 + ( 2 2 - f .06353 cm δ ring = ± 0

- f 1.54965 .06353 cm f ring ± δ ring = 2 ± 0 f − f 11. [5 pts] Calculate the spherical aberration as the percent difference . Also calculate δΔ f . f 00% △ = f small f − f ring small * 1 f 00% △ = (21.97259 cm ) (21.54965 cm )−(21.97259 cm ) * 1 f .924% △ = − 1 △ f δ = √ ( ) ) f ring δ f ring 2 + ( f small δ f small 2 △ f δ = √ ( ) ) 0.06353 21.54965 2 + ( 0.23684 21.97259 2 △ f .01117 or 1.1117% δ = 0 f △ f .924% .1117% △ ± δ = − 1 ± 1 12. [5 pts] Calculate the radius of curvature of the lens, R, and δ R . R = w 2 4( d − t ) + 4 d − t R = (7.2 cm ) 2 4(0.88 cm −0.34 cm ) + 4 (0.88 cm −0.34 cm ) 4.135 cm R = 2 R δ = √ δ (1 ) (δ ) w 2 4( d − t ) 2 w 2 + 1 16 − w 2 ( d − t ) 2 2 d 2 + δ t 2 R δ = √ (0.01) (1 ) (0.01 .01 ) (7.2) 2 4(0.88−0.34) 2 2 + 1 16 − (7.2) 2 (0.88−0.34) 2 2 2 + 0 2 R .62854 cm δ = ± 0

From the calculation of index of refraction n, , the results shows that n .55 .0144 n ± δ = 1 ± 0 our value is reasonable becasue it is greater than 1. Other Questions [8 pts]: 17. [4 pts] Suppose the image of an object is focused by a thin lens at the focal length of the lens (i.e. s ′ = f ). What is the object distance s ? The object distance s is Infinite. 18. [4 pts] How could we alter our test for spherical aberration to test a lens for chromatic aberration instead? We would need a smaller focal length so that the light rays do not bend as much. Virtual Experiment [15 pts] Instructions 1. Click here and visit the online tool website. 2. This simulation allows the user to interact with concave and convex lenses by changing several parameters. 3. You can change the position of the object by clicking the dot (or tip of the object arrow) with your mouse. Additionally, the sliders in the top-left allow you to adjust the focal length f and the size of the lens. 4. Adjust the focal length to 3.65 cm and the lens size also to 5 cm. You may need to zoom in to see 3 cm on the grid. 5. Using these preset values, you will work towards determining two properties of the lens: the index of refraction n and the radius of curvature R . 6. You can measure the object distance s and image distance s ′ using the grid in the simulation tool. Estimate digits after the decimal point (i.e. values between the grid lines) by eye. 7. As you fill out the data table, try your best to keep the height of the object constant. For this exercise, set the height of the object to 2cm. Questions 8. [4 pts] Fill the following table:

9. [5 pts] Use the data you collected in the previous question to find f x and f y described in the lab manual. From these, find the average focal length f . .083 s 1 = 1 12 = 0 .091 s 1 = 1 11 = 0 .100 s 1 = 1 10 = 0 .111 s 1 = 9 1 = 0 .125 s 1 = 8 1 = 0 .083 .091 .100 .111 .125 / 5 .102 1 S = 0 + 0 + 0 + 0 + 0 = 0 .190 1 s ′ = 1 5.25 = 0 s [cm] s [cm] ′ 12 5.25 cm 11 5.425 cm 10 5.75 cm 9 6.14 cm 8 6.72 cm

y .105 f = 1 9.512 = 0 ( ) f = 2 fy + fx = 2 1 1 B − B m ( ) .0835 f = 2 0.105+0.062 = 2 1 1 9.512 − 9.512 −0.5857 = 0 The average focal length f is 0.0835 cm 10. [4 pts] Find the radius of curvature. You will need to make measurements directly from the screen using the square grid to get the approximate thickness at the center of lens ‘d’. Use t=0.2 cm for the thickness of the lens at top. height 0 cm w = 1 center ( thickness center ) .72 cm d = 0 top ( thickness edge ) .2 cm t = 0 8.21 cm R = w 2 4( d − t ) + 4 d − t = (10) 2 4(0.72−0.2) + 4 0.72−0.2 = 4 The radius of the curvature is 48.21 cm 11. [2 pts] Find the index of refraction. Thin lens equation: → f = R 2( n −1) .0835 .0835 x 2( n ) 48.21 0 = 48.21 2( n −1) = 0 − 1 = → ( n ) 48.21 0.0835 = 2 − 1 77.37 ( n ) 5 = 2 − 1 .88 n − 1 = 2 .88 cm n = 3