Homework 4 Solution

Homework 4 Solution - Version PREVIEW – Homework 4 –...

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Unformatted text preview: Version PREVIEW – Homework 4 – Savrasov – (39825) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Concave Mirror 01 001 (part 1 of 2) 10.0 points An object is p = 33 . 5 cm in front of a concave mirror. Its real image height is 6 times larger than the object height. p h What is the location of the image? Correct answer: 201 cm. Explanation: 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 Let : M =- 6 and p = 33 . 5 cm . Let the location of the image be denoted as q . q p h h' f R Since the image is 6 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 =- 6. M = h ′ h =- q p =- 6 , so q =- M p =- (- 6) (33 . 5 cm) = 201 cm . 002 (part 2 of 2) 10.0 points What is the radius of curvature of the mirror? Correct answer: 57 . 4286 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 or R = 2 pq p + q = 2 (33 . 5 cm) (201 cm) 33 . 5 cm + 201 cm = 57 . 4286 cm . Convex Mirror 02 003 10.0 points A convex mirror has a focal length of 60 cm. What is the position of the resulting image if the image is erect and 5 times smaller than the object?...
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This note was uploaded on 01/10/2010 for the course PHY na taught by Professor Richardscalettar during the Fall '09 term at UC Davis.

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Homework 4 Solution - Version PREVIEW – Homework 4 –...

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