Physics 222 Spring 2009 Exam 3 Version D (852430)

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1 Physics 222 Spring 2009 Exam 3 Version D (852430) Question Instructions Be sure to answer every question. Follow the rules shown on the first page for filling in the Scantron form. Each problem is worth 10% of the exam. When you are finished, check with Dr. Mike or his TA to be sure you have finished the scantron correctly. 1. Question DetailsMirror Equation or Ray Trace (3) [750394] A concave mirror (R = 64 cm) is used to project a transparent slide onto a wall. The slide is located at a distance of 35.0 cm from the mirror, and a small flashlight shines light through the slide and onto the mirror. The setup is similar to that in Figure 25.19a. Figure 25.19a. The height of the object on the slide is 1.00 cm. What is the height of the image? (Make sure you give the proper sign in this case!) w*05*10.67 cm v*10* cm y*04*-0.09 cm x*03*2.21 cm z*02*0.09 cm Since the mirror is concave and spherical, it has a focal length of +R/2 = cm. Also, since the image is in front of the mirror, it is positive as well cm is the object distance since the slide is the object and it is positive too. The image is the projection on the wall, so the distance to the wall from the mirror must be the image distance. Thus, we can use the mirror equation to find d i = (1/f - 1/d o ) -1 = cm. The height of the object on the slide is h o and may be positive or negative depending on whether the object is considered to be right side up or upside down. Normally, we assume the object to be positive in height regardless of its actual orientation. This is why we use a positive number here for the object height. Using the linear magnification equation, we have h i = - h o d i / d o = cm. Note that this is negative because it is inverted and the object height is positive. 2. Question DetailsRay Trace or Thin-Lens Equation (1) [ ] Do a ray trace for the object and mirror shown here or use the number of blocks to do the thin-lens equation. Determine which of the following describes the image.

2 x*04*it is virtual, upright, and smaller than the object. z*02*it is virtual, inverted and smaller than the object. v*10*it is virtual, upright, and larger than the object. w*04*it is real, inverted, and larger than the object. y*02*it is real, upright, and smaller than the object. The ray trace is shown below.

3 3. Question DetailsRay Trace or Thin-Lens Equation (1) [ ] Do a ray trace for the object and mirror shown here or use the number of blocks to do the thin-lens equation. Determine which of the following describes the image.

4 z*02*it is virtual, inverted and larger than the object. x*04*it is virtual, upright, and larger than the object. w*04*it is real, inverted, and smaller than the object. v*10*it is virtual, upright, and smaller than the object. y*02*it is real, upright, and larger than the object.

5 4. Question DetailsReflection and Mirrors (1) [750389] Bailey-the-Wonderdog sees a reflection of herself in a mirror. When she does so she notices that the dog she sees in the image is smaller than she is. What is the shape of the mirror? v*10*convex or Concave would both give this result. y*05*convex z*01*there is no mirror that would give this result. x*01*flat w*05*concave There are cases where this is true for both types of mirror. For instance, if we combine the mirror equation with the magnification equation, we get m = 1 / (d o / f -1). Then, when f = 1 and d o = 3, we have m = -1/2. Also, when f = -3 and and d o = 3, we get m = 1/2. In both cases, the absolute value of the magnification is less than 1 and the image is smaller. There really is no way to tell without drawing the ray trace or doing the equations. Although both mirrors can give smaller images, the two types of mirrors do NOT both give larger images. Only one does this. Can you figure out which one?

6 5. Question DetailsTotal Internal Reflection (2) [749486] A ray of light falls on a rectangular glass block (n = 1.55) that is almost completely submerged in water (n = 1.33) as shown in Figure Figure In this case, we can find an angle for which total internal reflection occurs at the point P. Would total internal reflection occur at point P for this same value of if the water were replaced by another fluid with index of refraction n = 1.41? v*10*no w*04*yes x*03*it depends on the value of. y*03*it depends on the frequency of the light. z*01*there is no way to determine the answer. The angle necessary to get total internal reflection is given by the equation c = sin-1 (n 2 / n 1 ). Any angle less than this will also refract. For the case with water we have n 2 = 1.33 and for the other fluid n 2 = The value of n 1 = 1.44 in both cases. As you can see by chacking the equations, c is greater for the other fluid than it is for water. If to be c when the water is present, then the same less than the c for the fluid and refraction will occur. was just big enough for the angle at P gives us the same angle at P even with the fluid present. Thus, the new angle is 6. Question DetailsWhat can we see? (3) [750393] A person 1.62 m tall wants to be able to see her full image in a plane mirror. How far above the floor should the bottom of the mirror be placed so that she can see her feet, assuming that the top of the person's head is 20 cm above her eye level? y*06*0.81 m z*01*0.2 m w*05*1.01 m x*03*1.52 m v*10*0.71 m This problem is easy if you draw a diagram! The minimum height for the mirror must allow for light from her feet and the top of her head to reach her eyes. A ray diagram will easily show this to be 1/2 her height. If the top of her head is 20 cm above her eyes, then the light from her head as it goes to her eyes must hit the mirror 10 cm below her head (the position of the top of her head is her height.) Thus, the top of the mirror is 10 cm less than her height (1.52 m) and the bottom is half her height below that (0.810 m), or m. 7. Question DetailsMirror Equation (4) [750392]

7 Plane mirrors and convex mirrors form virtual images. With a plane mirror, the image may be infinitely far behind the mirror, depending on where the object is located in front of the mirror. For an object in front of a single convex mirror, what is the greatest distance behind the mirror at which the image can be found? Assume that R is the radius of the mirror z*03*4r/3 w*05*r v*10*r/2 y*03*3r/4 x*05*2r Look at the mirror equation, and solve for the image distance. d i = (1/f - 1/d o ) -1. If f is a negative number (as it is for a convex mirror) and d o is positive (as it always is for a single mirror or lens), then the largest possible distance of the image behind the mirror is when the object is infinitely far away. Thus, the image is at d i = -f = -R/2, which is R/2 behind the mirror. 8. Question DetailsMirror Equations and Magnification (3) [750390] Dr. Mike stands in front of a CONVEX mirror. He looks at the image of his eye. It is UPRIGHT and half the size of his real eye. If f is the focal point of the mirror, how far is he standing from the front of the mirror? x*03*3f/8 z*04*2f y*04*f/2 v*10*f w*03*2f/3 For this situation, the linear magnification is +1/2. Thus, we have +1/2 = - d i / d o or d i = - d o / 2. Using this in the mirror equation, we find f = [1 / d o - 2 / d o ] -1 = - d o. Therefore, d o = - f. 9. Question DetailsRay Trace or Mirror Equation (1) [ ] Do a ray trace for the object and mirror shown here or use the number of blocks to do the mirror equation. Determine which of the following describes the image.

8 y*02*it is real, upright, and smaller than the object. w*04*it is real, inverted, and larger than the object. v*10*it is virtual, upright, and larger than the object. z*02*it is virtual, inverted and smaller than the object. x*04*it is virtual, upright, and smaller than the object. The ray trace is shown below.

9 10. Question DetailsApparent Depth (1) [750391] A silver medallion is actually 2.27 cm beneath the surface of a calm pool. How far below the surface of the pool would someone see the medallion if they were viewing it from above the pool? (In other words, what is its apparent depth?) Note: The refractive index of water is y*03*0.44 cm w*01*3.02 cm v*10*1.70 cm z*00*2.27 cm x*04*0.59 cm The apparent depth is d' = d (1) / n water = 1.70 cm. Assignment Details Name (AID): Physics 222 Spring 2009 Exam 3 Version D (852430) Submissions Allowed: 100 Category: Exam Feedback Settings Before due date Nothing After due date Nothing

10 Code: Locked: No Author: DeAntonio, Michael ( mdeanton@nmsu.edu ) Last Saved: Apr 6, :07 PM MDT Permission: Protected Randomization: Person Which graded: Last