lab 6 physics

Learning Goal: To learn and practice the geometry skills necessary for complex reflection setups. The law of reflection has the very simple form - Oreflected, Oincident - where incident is the angle between the normal and the incident ray and reflected is the angle between the normal and the reflected ray. Although the law itself is easy to use, many realistic situations involve successive reflections from multiple surfaces. The law of reflection does not become any more complicated in such cases, but the geometry of the rays does become complicated. Consider the case of light shining onto a mirror, which is attached to another mirror at some angle a, as shown in the figure (Figure 1). In this problem, we will find the angle at which light leaves the arrangement of two mirrors. Part A If the light strikes the first mirror at an angle 01, what is the reflected angle 02? Express your answer in terms of 0₁. VTI ΑΣΦ 0₂ = Submit Request Answer ?

Normal line Li= P S Z Ls Normal line B Le = 30° ● - Dot represents a pin Q Step 9 C Step 10 Calculations: 1. Using Snell's Law of Refraction calculate the index of refraction for the piece of glass from the experiment. 2. Calculate the speed of light inside the glass. 3. Use your index of refraction to find the angle of refraction that would be produced if light entered your glass at an angle of incidence of 60 degrees. Show all work.

explain the steps to find the answer, which equation you are using and why you are using it. Also a small reflection about the problem.  Thanks.

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What are the factors that affects the refraction of light? Why is it important to understand the speed of light in different mediums? Site 2 industrial applications of this concept.

1. Refractive Material The following are top-view diagrams of solid cylinders and cubes. Assume that light travels more slowly through the objects than through the surrounding medium. Each diagram shows a path for light that is not qualitatively correct; there is at least one flaw, perhaps more, in each diagram. Identify all flaws. Explain your reasoning. a. b. d.

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Please only answer question 9. I included question 8 as well to give the background of the question. Please give a detailed explanation for why the total internal reflection occurs when the incident angle is greater than the critical angle. Please explain each step carefully with good reasoning.

2 and 3 answer pls

7. What happens when the incident angle is greater than the critical angle? Please describe least two complete sentences. 8. This phenomenon is called Total Internal Reflection (TIR). What do you think would be some practical purposes of TIR in modern technology? Please name at least one example and explain your answer in at least two complete sentences.

Help physics HW Using Excel create a plot graph of reflected angle (y-axis) vs. incident angle (x-axis).

This question is related to to the experiment in which we find angle of prism and angle if minimum deviation using a spectrometer. Explain the attached statement .

Problem Description A laser beam travels from air (n=1) into glass (n=1.5) and then into gelatin. The incident ray makes a 58.0° angle with the normal as it enters the glass and a 36.4° angle with the normal as it enters the gelatin. Instructions In a neat and organized fashion, write out a solution which includes the following: 1. A sketch of the physical situation with all given physical quantities clearly labeled. 2. Draw a ray diagram showing all reflected and refracted rays in this situation and all angles clearly labeled. 3. Determine (a) the angle the refracted ray makes with the normal in the glass and (b) the index of refraction of the gelatin. Clearly show all steps, starting from generalized equations. Explain your mathematical work in words. Your explanation should cover both what you did and the thought process behind why you did that. 4. Evaluate your answer to determine whether it is reasonable or not. Consider all aspects of your answer (the numerical value, sign, and…

Part A X rays with a wavelength of 0.19 nm undergo first-order diffraction from a crystal at a 50° angle of incidence. At what angle does first-order diffraction occur for x rays with a wavelength of 0.15 nm ? Express your answer using two significant figures. 0 = ΜΕ ΑΣΦ Submit Request Answer Provide Feedback ? °

air crystal Consider 2 rays of light inside of a clear crystal. Both rays are incident upon the crystal/air boundary at the same point with the angles indicated, and ray A is internally reflected. Here, take a = B= 30 degrees. Choose all of the following statements that are true: Select one or more: a. In order that the ray A is internally reflected, the index of refraction has to be anything greater than 2 9 b. The speed of the rays in the crystal is a bit slower than the speed of light O c. The speed of the rays in the crystal is a bit faster than the speed of light O d. Since A is internally reflected, B also must be internally reflected. e. The speed of the rays in the crystal is equal to the speed of light f. In order that the ray A is internally reflected, the index of refraction has to be anything greater than 1.15

The H & D response curves for two kinds of film in a film camera are shown in the figure. Which film would be best for each of the following situations? In each case tell which film characteristic (speed, contrast, or latitude) is the important one.a) Photographing in a situation where there is very little light.b) Photographing in a situation where there is a large variation in light level. c.) In above H & D curve, which film (between A and B) is faster?

Please explain steps, thank you

Assessment 2 Directions: Write ULTRA if the statement is correct, and VIOLET if not. If the statement is incorrect, state reason or appropriately correct the statement. 1. Too much exposure to UV rays may increase the risk for skin cancer and cataracts. 2. Sunscreens protect our skin so we can stay under the sun all day. 3. The higher the Sun Protection Factor (SPF) in a sunscreen, the higher the protection. 4. Sunglasses have UV coating that help block harmful rays from the sun. 5. It is good to avoid the sun entirely.

6) Hi, Can you solve the question 6 and make explanation please? and if you are gonna use handwrite please make sure that it is readable thank you.  I know that you cant solve multiple options so can you solve the option A and B please

2 Problem 2 Assume two adjacent interstellar gas clouds with known thickness dị and d2 found in front of an astronomical light source. Suppose that this source emits a beam of radiation with known specific intensity 1, as it enters the first cloud of interstellar gas. Calculate the specific intensity I, immediately after this beam has traversed the two clouds of gas. Assume that the densities (p1 and p2) and the opacities (k,1 and ky,2) are constant within the clouds and are known quantities and that these clouds possess no emissivity. (Note the relationship between the opacity and absorption coefficient a, = pk,. See Rybicki and Lightman eq. 1.22.) %3D

d. Diameter of the Sun A Standing on the beach at sunset, you extend the tip of your finger at your full arm length from your face, covering the Sun. Upon moving your finger around, you find that only about half of its width is needed to completely cover the Sun's diameter. You measure your finger width to be 0.5 inches. You know your arm length to be 28.0 inches. You have be told that the Sun is approximately 93 million miles away. Use this information to determine the approximate diameter of the Sun, filling in the table below with the proper quantities measure in meters. 1= ½ finger g= cye level height d = object A = angle from eye level to top of eye level object f= object height from x = Arm length h = Diameter width distance of the Sun What is the diameter of the Sun in Miles? The actual diameter of the Sun more like 865,000 miles. What is the %-Error in your calculated value?

Figure 1: Schematical view of an absorber layer and back reflector NAir=1 R L Nabs=3 Oc Oc Lambertian scatterer The absorber layer shown in Figure 1 has a Lambertian scatterer at the back side. The absorber layer thickness is equal to D = 1 [µm]. The critical angle for light propagating from the absorber layer to air is indicated in the figure by Oc. A) What fraction of Ig is scattered into the critical angle Oc,Abs-Air ? Give a percentage %. For the angles smaller than the critical angle, light can escape the absorber layer. The area under which light can escape the absorber layer is called the escape cone, indicated by the red area in Figure 1. B) Calculate the fraction of angles under which light can escape the absorber layer. Give a percentage. (Hint: the fraction of angles is equal to the ratio of the area of the escape cone, to the surface area of the full hemisphere with radius L into which light can be emitted.)

The isosbestic point is defined by the intersection of all spectra at the same point where; A. ABS = &,bC*EbC, B. ABS = &,bC, +E,bC, %3D C. &bC =8,bC, D. & =E2 Select answer Save O developed by nextech EGO

need help with this, thank you

can you solve step by step

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Mystery Material Material 1 Light passes from Material 1 into a new, Mystery Material. The angle of incidence of the light on the interface between Material 1 and the Mystery Material is 0, n, in mystery 45°. %3D a) Draw 0, in the diagram to the right. b) Draw the refracted ray and angle of the light if Mystery Material has a lower index of refraction compared to Material 1 Mystery Material Material 1. n, mystery Nmystery n1 Note: no numbers are necessary, just the general path and shape of the ray and angle.