MIDTERM 4 - Version 005/AAABB – midterm 04 – Turner –...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Version 005/AAABB – midterm 04 – Turner – (59130) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points The magnitude of the radius of a spherical mirror is | R | = a . An object is placed in front of this spherical mirror. It forms a virtual image. The magnification is M = 1 2 . The mirror is 1. convergent 2. divergent correct Explanation: Since there is a reduced vertical image, it must be a convex (divergent) mirror. 002 (part 2 of 2) 10.0 points The object distance is 1. 1 4 a 2. 1 2 a correct 3. 3 2 a 4. 5 2 a 5. 1 3 a 6. a 7. 3 a 8. 2 3 a 9. 2 a 10. 3 4 a Explanation: Basic Concepts: 1 p + 1 q = 1 f = 2 R m = h ′ h = − q p Concave Mirror f > ∞ >p> f f <q < ∞ >m> −∞ f >p> −∞ <q < ∞ >m> 1 Convex Mirror > f ∞ >p> f <q < <m< 1 Solution: Since there is a reduced virtual image, it must correspond to convex mirror, or R = −| R | . Also, virtual image implies p q = − p | q | = − 1 | M | = − 2 . Hence, 1 p + 1 q = 2 R 1 p parenleftbigg 1 − 1 | M | parenrightbigg = − 2 | R | 1 p (1 − 2) = − 2 a − 1 p = − 2 a . Notice: The last equality confirms the fact that our choice of R = −| R | is correct. ⇒ p = a 2 003 10.0 points The reflecting surfaces of two intersecting flat mirrors are at an angle of 65 ◦ , as shown in the figure. A light ray strikes the horizontal mirror at an angle of 48 ◦ with respect to the mirror’s surface. 65 ◦ 48 ◦ φ Figure is not drawn to scale. Calculate the angle φ . Version 005/AAABB – midterm 04 – Turner – (59130) 2 1. 68.0 2. 66.0 3. 62.0 4. 70.0 5. 52.0 6. 50.0 7. 48.0 8. 76.0 9. 56.0 10. 64.0 Correct answer: 50 ◦ . Explanation: Basic Concept: θ incident = θ reflected Solution: θ 1 θ 2 φ θ Figure is to scale. The sum of the angles in a triangle is 180 ◦ . In the triangle on the left we have angles θ , 180 ◦ − θ 1 2 , and 180 ◦ − θ 2 2 , so 180 ◦ = θ + 180 ◦ − θ 1 2 + 180 ◦ − θ 2 2 , or θ 1 + θ 2 = 2 θ . (1) In the triangle on the right we have angles θ 1 , θ 2 , and φ. 180 ◦ = θ 1 + θ 2 + φ, so θ 1 + θ 2 = 180 ◦ − φ. (2) Combining Eq. 1 and 2, we have φ = 180 ◦ − 2 θ = 180 ◦ − 2 (65 ◦ ) = 50 ◦ . As a matter of interest, in the upper-half of the figure the angles (clockwise) in the triangles from left to right are 42 ◦ , 42 ◦ , and 96 ◦ ; 84 ◦ , 25 ◦ , and 71 ◦ ; 109 ◦ , 23 ◦ , and 48 ◦ ; 132 ◦ , 23 ◦ , and 25 ◦ ; and in the lower-half of the figure the angles (counter-clockwise) in the triangles from left to right are 23 ◦ , 23 ◦ , and 134 ◦ ; 46 ◦ , 25 ◦ , and 109 ◦ ; 71 ◦ , 42 ◦ , and 67 ◦ ; 113 ◦ , 42 ◦ , and 25 ◦ ....
View Full Document

This note was uploaded on 02/17/2010 for the course PHY 59130 taught by Professor Turner during the Fall '09 term at University of Texas.

Page1 / 10

MIDTERM 4 - Version 005/AAABB – midterm 04 – Turner –...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online