This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: nguyen (jmn727) homework 36 Turner (59070) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points An object located 34 . 3 cm in front of a lens forms an image on a screen 9 . 41 cm behind the lens. Find the focal length of the lens. Correct answer: 7 . 38419 cm. Explanation: Basic Concepts: 1 p + 1 q = 1 f m = h h = q p Converging Lens f > >p> f f <q < >m> f >p> <q < >m> 1 Diverging Lens f < >p> f <q < <m< 1 Solution: The formula relating the focal length to the image distance s and the object distance s is 1 s + 1 s = 1 f so f = s s s + s = (34 . 3 cm) (9 . 41 cm) (34 . 3 cm) + (9 . 41 cm) = 7 . 38419 cm . 002 (part 2 of 2) 10.0 points What is the magnification of the object? Correct answer: . 274344. Explanation: Magnification is M = s s = (9 . 41 cm) (34 . 3 cm) = . 274344 . 003 (part 1 of 2) 10.0 points A convergent lens has a focal length of 25 . 1 cm . The object distance is 11 . 7 cm . f f q h p h Scale: 10 cm = Find the distance of the image from the center of the lens. Correct answer: 21 . 9157 cm. Explanation: 1 p + 1 q = 1 f M = h h = q p Convergent Lens f > f > p> <q < >M > 1 Note: The focal length for a convergent lens is positive, f = 25 . 1 cm. Solution: Substituting these values into the lens equation q = 1 1 f 1 p = 1 1 (25 . 1 cm) 1 (11 . 7 cm) = 21 . 9157 cm  q  = 21 . 9157 cm ....
View Full
Document
 Spring '08
 Turner
 Work

Click to edit the document details