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OPTI 201R Homework 3 Homework Is Due in Class. Do All Problems And OPTI 201R Homework 3 Homework is due in class. Do all problems and show your work. Credit is not given for answers only. You are welcome to work together, but be sure your homework is your work. 1. The figure below shows a Cartesian oval which is a refracting surface where the optical path length (OPL) between point S and point P is the same regardless of where a ray leaving point S intersects the surface. a) What is the OPL for rays following the path SVP? The indices of refraction are n0 and n1 on either side of the surface. OPL ∑ nL, so the OPL for the path SVP is given by b) Show that the OPL for rays following the path SAP is given by 22 2 2 0 101 yxssnyxnOPL , where sx 0 . The OPL for the path SAP is given by . However, we need to solve for and in terms of , , y, and ∆, given that x= ∆. In order to solve for , we look at the right triangle whose hypotenuse is , and use the Pythagorean Theorem which states that , where a and b are the sides of the triangle and c is the length of the hypotenuse. The hypotenuse is then given by ∆ , and when making the substitution x= ∆, we get . Next we use the same process to find . Where the sides are given by ∆ and y. The horizontal length ( ∆) can be rewritten to become equal to , which leads to the length . Finally, we just plug in our equations for and into our starting equation and we successfully obtain the equation that we were first given. 2. For the dove prism shown below, what does the distance AB need to be so that a ray entering the prism at point A with height y will emerge from the prism at point B with a height y. Assume the prism’s refractive index is n = 1.5. The first step is to find the angle of incidence on the prism and apply Snell’s law to determine the angle of the ray within the glass. The angle of incidence is 45 degrees. Applying Snell’s law results in an angle of =28.1255 degrees. However, the angle we need is 45‐, or 16.8745 degrees. Using trig we can determine that the distance AC is given by . This leads to AC=3.2967*y, so the total distance AB then becomes equal to 2*AC, which means that AB=6.5933*y. See figure below for where the point C is located and where is located. 3. Figure (a) below shows a 3D view of how the letter R emerges from a right prism. The parity is switched since there is a single reflection (recall the eye looks back along the ray to see the R with switched parity). Figure (b) below shows a cross‐sectional view of a Taylor prism and how an incident ray moves through the prism. Figure (c) below shows a 3D view of the Taylor prism. Sketch on this figure the orientation of the letter R as it emerges from the prism. Is parity switched in this case? The parity is switched since there are three reflection (i.e. odd number of reflections). 4. The prism below consists of two parallel sides with a thickness of 2B. The bottom side of the prism has an acute angle of 60°. The upper side of the prism has an acute angle of 30°, as well as a roof. A ray is incident from the left and strikes the bottom face at a point that bisects the surface. Assuming the refractive index is high enough for total internal reflection at each of the surface reflections for this ray, do the following. (a) Draw a neat diagram of how this ray propagates through the prism and where it emerges from the prism. Show all of the surface normals at each reflection and label with the angles of incidence and reflection, with the proper sign convention. (b) What is the angle between the incident and emerging ray? The incident and emerging rays are parallel. (c) Is the parity conserved for this prism? There are four total reflections: one from the bottom face, one from the right face and two from the roof on the top face. Since the number of reflections is even, parity is conserved. (d) How much higher is the emerging ray relative to the incident ray? Since the incident ray strikes the bisector of the bottom surface, the base of the red triangle needs to have a width B. Based on the angle of incidence on the right face, the upper angle of the red triangle must be 30°. This in turn means the height of the red triangle is B / tan 30° = 3 B. From the symmetry of the problem, the height between the emerging and incident rays is 2 3 B. (e) If a letter R (oriented in and out of the plane of the paper) is placed at the location shown in the drawing and we look back along the emerging ray, draw the orientation and parity of the emerging letter R. Since the parity is conserved (see part (c)), the emerging letter R should have right‐handed parity. We just need to determine the orientation of the emerging letter. The reflections from the bottom and right faces invert the letter and the reflections from roof reverts the letter, so the emerging letter appears to be rotated 180° from the original. .
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