![]() ![]() Thus, the angle of refraction should be smaller than the angle of refraction. In this problem, the light ray is traveling from a less optically dense or fast medium (air) into a more optically dense or slow medium (water), and so the light ray should refract towards the normal - FST. When finished, it is always a wise idea to apply the FST and SFA principles as a check of your numerical answer. Proper algebra yields the answer of 36.2 degrees for the angle of refraction. Now list the relevant equation (Snell's Law), substitute known values into the equation, and perform the proper algebraic steps to solve for the unknown. N r = 1.333 (from table) Θ i = 52 degrees The solution to this problem begins like any problem: a diagram is constructed to assist in the visualization of the physical situation, the known values are listed, and the unknown value (desired quantity) is identified. Refer to the table of indices of refraction if necessary. Determine the angle of refraction of the light ray. In this part of Lesson 2, we will investigate several of the types of problems that you will have to solve, and learn the task of tracing the refracted ray if given the incident ray and the indices of refraction.Įxample Problem A A ray of light in air is approaching the boundary with water at an angle of 52 degrees. If any three of the four variables in the equation are known, the fourth variable can be predicted if appropriate problem-solving skills are employed. N r = index of refraction of the refractive mediumĪs with any equation in physics, the Snell's Law equation is valued for its predictive ability. N i = index of refraction of the incident medium Where Θ i ("theta i") = angle of incidence The equation is known as the Snell's Law equation and is expressed as follows. Part1\Geometrical Set.1\Plane.1\Activity true:Īs you probably know, a 2D element in CATIA can be created on a planar surface or plane so you need to think in different terms your problem.remember that in CATIA you have also different views (iso, front view a.s.o) and simple command "Fit All In" will change how the elements will be shown on your screen.In a previous part of Lesson 2, we learned about a mathematical equation relating the two angles (angles of incidence and refraction) and the indices of refraction of the two materials on each side of the boundary. Part1\Geometrical Set.1\Plane.1\Offset 0mm: Part1\Absolute Axis System\Activity true: If you will put 4 rectangles it will depends where you will create.īellow is a list of parameters for a plane created in a CATPart. Hope you notice that any CATPart has an Axis System and xy, yz, zx planes so you have to refer to those and not to the screen. My question would be, how can I convert the 2D rectangle coordinates to 3D coordinates, so I can move them to the right location in 3D space. ![]() And the 3D plane which is parallel to the screen can be computed. Move the second rectangle below the first rectangle, continue the for loop till all the rectangles are placed along the border of the screen.Īnother thing to notice here is, the dotted rectangles can be located along the xy plane or yz plane or xz plane. Get the 2 3D vectors(Up direction, right direction of the screen)Ĭompute the coordinates of corners of the screen by moving the 3D center point of the screen, by a distance equal to half the width, half the height of the screen, along the 2 vectors.Įquate the top right coordinate of the 2D rectangle to top right 3D coordinate of the screen.(which is giving the wrong output) I get the center point of the screen, which is a 3D point. I could not place them at the right location and I need your help. The problem here is, the rectangles are 2D, and the CATIA screen in a 3D space. The dotted rectangles shows the boxes in CATIA, and the solid rectangles are where I need to place them. I have a CATIA software in which there are a few rectangle boxes that I need to place along the border of the screen, as shown in the image. I have a problem regarding placing the 2D rectangles in a 3D space. ![]()
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