homework 7 - montes(dam2772 – Homework7 – shih...

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Unformatted text preview: montes (dam2772) – Homework7 – shih – (56310) 1 This print-out should have 20 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 Hint: Construct a ray diagram. Given: A real object is located at “ p 1 = 13 7 f ” to the left of a convergent lens with a focal length f as shown in the figure below. 13 7 f f ( × f ) The image distance q 1 to the right of the lens is 1. q 1 = 13 6 f . correct 2. q 1 = 11 6 f . 3. q 1 = 5 2 f . 4. q 1 = 15 4 f . 5. q 1 = 13 5 f . 6. q 1 = 15 8 f . 7. q 1 = 9 4 f . 8. q 1 = 13 7 f . 9. q 1 = 11 4 f . 10. q 1 = 13 4 f . 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 Convergent (convex) lenses f > 0 have real images q > 0 when the object ∞ > p > f . 13 7 f f 13 6 ( × f ) Basic Concepts: 1 p + 1 q = 1 f . Solution: 1 q 1 =- 1 p 1 + 1 f =- 7 13 f + 1 f = 13- 7 13 f = 6 13 f q 1 = 13 6 f . 002 (part 2 of 2) 10.0 points A second convergent lens with the same focal length f is placed behind the first lens, as shown in the figure below (the first lens has the lighter image). montes (dam2772) – Homework7 – shih – (56310) 2 13 7 5 6 f f f f ( × f ) The second image location q ′ 2 , measured with respect to the second lens, is 1. q ′ 2 = 1 2 f . 2. q ′ 2 = 3 4 f . 3. q ′ 2 = 2 3 f . 4. q ′ 2 = 5 7 f . 5. q ′ 2 = 9 16 f . 6. q ′ 2 = 5 8 f . 7. q ′ 2 = 8 13 f . 8. q ′ 2 = 4 7 f . correct 9. q ′ 2 = 3 5 f . 10. q ′ 2 = 7 12 f . Explanation: 59 42 f 13 7 4 3 f 5 6 13 6 4 7 f f ( × f ) Given: Distance between lenses is d = 5 6 f , and from Part 2, we have q 1 = 13 6 f . Solution: q 1 > d and for the second lens p 2 =- q 1 + d = parenleftbigg- 13 6 + 5 6 parenrightbigg f =- 4 3 f . Note: Since the object for the second lens is behind the lens it is virtual and therefore p 2 < 0. Using the lens equation for the second lens, we have 1 q 2 =- 1 p 2 + 1 f = 1 q 1- d + 1 f = 1 13 6 f- 5 6 f + 1 f = 6 8 f + 8 8 f = 6 + 8 8 f = 14 8 f q 2 = 4 7 f . The final image due to the second lens is real and to the right of the second lens. As can be seen from the figure, the final image due to the second lens relative to the first lens is q ′ 2 = 5 6 f + 4 7 f = 59 42 f . keywords: 003 (part 1 of 2) 10.0 points An cylindrical opaque drinking glass has a diameter 4 . 4 cm and height h , as shown in the figure. An observer’s eye is placed as shown (the observer is just barely looking over the montes (dam2772) – Homework7 – shih – (56310) 3 rim of the glass). When empty, the observer can just barely see the edge of the bottom of the glass. When filled to the brim with a transparent liquid, the observer can just barely see the center of the bottom of the glass....
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homework 7 - montes(dam2772 – Homework7 – shih...

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