Mastering Ray Tracing and Mirror Equations in Physics 132
School
University of Illinois, Chicago**We aren't endorsed by this school
Course
PHYSICS 132
Subject
Physics
Date
Dec 10, 2024
Pages
5
Uploaded by ColonelEmu4690
Physics 132 – Fall 2024Group Problems: Chapter 13Learning Goals After working through this worksheet, students will be able to: 1.Use ray-tracing to graphically find the location, orientation, and magnification of a final image. 2.Use the mirror equation to numerically find the same properties of a final image. 3.Conceptually describe how light is refracted at a plane surface. 4.Compute the incident or refracted angle that a ray of light makes at a plane surface, given the indices of refraction on either side of the surface. 5.Determine the position and magnification of an image formed by refraction at a plane surface. Problem 1:Ray tracing for concavemirrors A.Let’s first place the object beyondthe focal point (meaning ࠵?!> ࠵?) of a concave mirror (see picture). 1.Draw all fourspecial rays. Use them to determine the location of the image on the picture. 2.Compared to the object, is the image a.upright or inverted? b.real or virtual? c.larger or smaller? 3.The mirror focal length is ࠵? = +80 cmin the scaled downpicture. What is then the location of the object and of your image? (Hint:use the horizontal scale along the axis shown in the picture) 4.Use the mirror equation and the magnification relation: 1࠵?!+1࠵?"=1࠵?and࠵? = −࠵?"࠵?!To determine the location of the image and its magnification. B.Let’s now consider placing the object betweenthe focal point and the mirror, as pictured. 1.Pick 2 (or more) special rays, draw them, and determine the location of the image on the picture. 2.Compared to the object, is the image a.upright or inverted? b.real or virtual? c.larger or smaller? In a ray diagram for a mirror, drawing any tworays from a point on the object reflecting off the mirror determines the location of its image (they intersect at the location of the image). There are however four specialrays you can choose from, consistent with the law of reflection, that are easy to draw. All these special rays represent light emitted by the top of the object (draw them all starting at its top). For a concavemirror they are: 1)A ray that travels parallelto the optical axis, reflecting through its focal point. 2)A ray that travels throughthe focalpoint, reflecting parallel to the optical axis. 3)A ray that travels through the centerof curvature of the mirror, reflecting along the same line that it came from. 4)A ray that hits the vertexof the mirror, reflecting symmetrically about the optical axis. Inverted Real larger The location of the focal length sets the scale. The object must be located at ࠵?!= +120 cmand the image around ࠵?"= +240 cm1࠵?!+1࠵?"=1࠵?1120 cm+1࠵?"=180 cm࠵?"= +240 cmUpright Virtual larger
2 3.Pick 2 (or more) special rays, draw them, and determine the location of the image on the picture. 4.Given the mirror focal length of ࠵? = +80 cmuse the horizontal scale along the axis in the picture to determine the location of the object and of the image. 5.Use the mirror equation and the magnification relation to determine the image location and magnification, compare with your location found using ray-tracing. Problem 2:Ray tracing for convexmirrors A.Refer to the figure depicting an object (green arrow) placed in front of a convex mirror. 1.Use ray tracing with two (or more) special rays to find the location of the image (see picture). 2.Compared to the object, is the image a.upright or inverted? b.real or virtual? c.larger or smaller? 3.The mirror focal length is ࠵? = −60 cmin the scaled downpicture. Use the horizontal scale along the axis in the picture to determine the location of the object and of the image. 4.Use the mirror equation and the magnification relation to determine the image location and magnification, compare with your location found using ray-tracing. For a convexmirror we still have four special rays to choose from as before, but the first three change a bit: 1*)A ray that travels parallelto the optical axis, reflecting as ifcoming from the focal point. 2*)A ray that travels towardsthe focalpoint, reflecting parallel to the optical axis. 3*)A ray that travels towardsthe centerof curvature of the mirror, reflecting along the same line that it came from. 4)A ray that hits the vertexof the mirror, reflecting symmetrically about the optical axis. Upright Virtual smaller According to the scale the object must be located at ࠵?!= +60 cmand the image around ࠵?"= −35 cmor so.1࠵?!+1࠵?"=1࠵?160 cm+1࠵?"= +1−60 cm࠵?"= −30 cmNotice that in this case the mirror equation is a bit off. The origin of this discrepancy is the fact that in reality only parabolicalmirrors have a focal point. You can see this in the image tracing: the rays don’t all seem to be intersecting at a single location, a fact described as aberration.With our spherical mirror this image will be a bit distorted and out of focus.According to the scale the object must be located at ࠵?!= +40 cmand the image around ࠵?"= −80 cm1࠵?!+1࠵?"=1࠵?140 cm+1࠵?"= +180 cm࠵?"= −80 cm
3 Problem 3:Refraction at a plane interface and Snell’s law Here you will identify angles of incidence and apply Snell’s law to determine the refracted rays: ࠵?"sin(࠵?") = ࠵?#sin(࠵?#).A.In all cases below a ray is shown in air (࠵? = 1) incident on another material (grey, with ࠵? > 1). 1.For each case: a.draw the normal line b.indicate the angle of incidence c.sketch the refracted ray and indicate the angle of refraction. B.The incident rays now travel inside the high index of refraction material as shown. The incident angle is largerthan the critical angle only in case 6, and smallerthan it for other cases. 1.Which rays will experience total internal reflection? Explain your reasoning. 2.For each case which do not experience total internal reflection: a.draw the normal line b.indicate the angle of incidence c.sketch the refracted ray and indicate the angle of refraction. C.In case 10, the incident ray comes in at 60°from the surface, in air. 1.What is the angle of incidence? Draw it on the diagram. 2.We want the refracted angle to be ࠵?$= 10°. What index of refraction should the material have? Case 10θ=60°Case 6Case 7Case 8Case 9Case 1Case 2Case 3Case 4Case 5The angle of incidence is the angle between the ray and the normal, which is the complement of the 60°angle shown; thus, ࠵?"= 90° − 60° = 30°.Only case 6 will, since the incident angle is larger than the critical angle. There is then no refraction, only reflection.Using Snell’s law:࠵?#sin ࠵?"= ࠵?$sin ࠵?%⇒ ࠵?&"%sin 30° = ࠵?$sin 10°Plugging in ࠵?#= ࠵?&"%= 1and solving for ࠵?$gives ࠵?$=࠵?&"%sin(30°)sin(10°)= 2.879࠵?’࠵?!࠵?"࠵?!࠵?"࠵?"࠵?!࠵?!࠵?"࠵?!= 0࠵?"= 0࠵?!= 0࠵?"= 0࠵?"࠵?!࠵?!࠵?"
Problem 1. Part BProblem 1. Part ACfCf
Problem 2. Part ACf