# solution8_pdf - white(rw22679 Hw08 Ross(10219 This...

This preview shows pages 1–2. Sign up to view the full content.

white (rw22679) – Hw08 – Ross – (10219) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The magnification produced by a converging lens is found to be 2 . 6 for an object placed 22 cm from the lens. What is the focal length of the lens? Correct answer: 35 . 75 cm. Explanation: Let : M = 2 . 6 and p = 22 cm . 1 p + 1 q = 1 f M = h h = - q p Converging Lens f > 0 >p> f f <q < 0 >m> -∞ f >p> 0 -∞ <q < 0 >m> 1 From the magnification q = - M p so 1 f = 1 p - 1 M p = M - 1 M p = 2 . 6 - 1 (2 . 6)(22 cm) = 0 . 027972 cm 1 f = 1 0 . 027972 cm 1 = 35 . 75 cm . 002 (part 1 of 2) 10.0 points A certain LCD projector contains a single thin lens. An object 31 . 3 mm high is to be projected so that its upright image fills a screen 2 . 2 m high. The object-to-screen distance is 1 . 48 m. Find the focal length of the projection lens. Correct answer: 20 . 4698 mm. Explanation: Given : h = 31 . 3 mm = 0 . 0313 m , h = - 2 . 2 m , and L = 1 . 48 m . It is desired to form a magnified, real image on the screen using a single thin lens. To do this, a converging lens must be used and the image will be inverted. The magnification then gives M = h h = - q p q = - h h p = - - 2 . 2 m 0 . 0313 m p = 70 . 2875 p. Also we know that L = p + q = 1 . 48 m, so p + 70 . 2875 p = 1 . 48 m p = 1 . 48 m (1 + 70 . 2875) · 1000 mm 1 m = 20 . 761 mm . The lens equation gives 1 p + 1 q = 1 f 1 p + 1 70 . 2875 p = 1 f f = 70 . 2875 p 70 . 2875 + 1 = (70 . 2875) (20 . 761 mm) 70 . 2875 + 1 = 20 . 4698 mm . 003 (part 2 of 2) 10.0 points How far from the object should the lens of the projector be placed in order to form the image on the screen?

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern