{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Lemuel physics 2 2 - Wong Lemuel Homework 2 Due Feb 2 2007...

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

View Full Document Right Arrow Icon
Wong, Lemuel – Homework 2 – Due: Feb 2 2007, 11:00 pm – Inst: Matt Mackie 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. Note: HW is due 11pm central time, which is midnight out time. 001 (part 1 of 1) 10 points A concave mirror with a radius of curvature of 1 . 2 m is illuminated by a candle located on the symmetry axis 3 . 2 m from the mirror. Where is the image of the candle? Correct answer: 0 . 738462 m. Explanation: 1 p + 1 q = 1 f = 2 R m = h 0 h = - q p Concave Mirror f > 0 > p > f f < q < 0 > m > -∞ f > p > 0 -∞ < q < 0 > m > 1 Let : f = 0 . 6 m and p = 3 . 2 m . From the mirror equation 1 p + 1 q = 1 f = 2 R f = R 2 q = 1 f - 1 p - 1 = 1 (0 . 6 m) - 1 (3 . 2 m) - 1 = 0 . 738462 m . keywords: 002 (part 1 of 2) 10 points An object is p = 38 . 4 cm in front of a concave mirror. Its real image height is 8 times larger than the object height. p h What is the location of the image? Correct answer: 307 . 2 cm. Explanation: 1 p + 1 q = 1 f = 2 R m = h 0 h = - q p Concave Mirror f > 0 > p > f f < q < 0 > m > -∞ f > p > 0 -∞ < q < 0 > m > 1 Let : M = - 8 and p = 38 . 4 cm . Let the location of the image be denoted as q . q p h h' f R Since the image is 8 times larger than the object, and is a real image, as stated in the problem ( p > 0 and q > 0), the magnification must be negative; i.e. , M = - 8. M = h 0 h = - q p = - 8 , so q = - M p = - ( - 8) (38 . 4 cm) = 307 . 2 cm . 003 (part 2 of 2) 10 points What is the radius of curvature of the mirror? Correct answer: 68 . 2667 cm. Explanation: Now that we have q and p , we can find the radius of the curvature via the relationship 1 p + 1 q = 2 R
Image of page 1

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

View Full Document Right Arrow Icon
Wong, Lemuel – Homework 2 – Due: Feb 2 2007, 11:00 pm – Inst: Matt Mackie 2 or R = 2 p q p + q = 2 (38 . 4 cm) (307 . 2 cm) 38 . 4 cm + 307 . 2 cm = 68 . 2667 cm . keywords: 004 (part 1 of 1) 10 points A concave mirror has a focal length of 23 cm. What is the magnification if the object’s distance is 274 cm? Correct answer: - 0 . 0916335 . Explanation: Basic Concepts: 1 p + 1 q = 1 f = 2 R m = h 0 h = - q p Concave Mirror f > 0 > p > f f < q < 0 > m > -∞ f > p > 0 -∞ < q < 0 > m > 1 Solution: Let q be the image distance and o the object distance. From the mirror equation 1 o + 1 q = 1 f m = - q o Combining the two equations yields m = f f - o = - 0 . 0916335 . keywords: 005 (part 1 of 1) 10 points A convex mirror has a focal length of - 24 cm. What is the object distance if the image distance is - 13 cm? Correct answer: 28 . 3636 cm. Explanation: 1 p + 1 q = 1 f = 2 R m = h 0 h = - q p Convex Mirror 0 > f > p > 0 f < q < 0 0 < m < 1 Let : f = - 24 cm and i = - 13 cm . From the mirror equation 1 o + 1 i = 1 f Solving for o yields o = i f i - f = ( - 13 cm) ( - 24 cm) ( - 13 cm) - ( - 24 cm) = 28 . 3636 cm .
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern