Chapter 17 - Light and Image Formation
Chapter 17 - Light and Image Formation
Chapter 17 - Light and Image Formation
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Announcements Reflection <strong>and</strong> Refraction <strong>Image</strong> <strong>Formation</strong> Final QuestionsQuestionA diverging lens has a focal length of 10.0 cm. An object is placed 30.0 cm from the lens. Find the image distance<strong>and</strong> describe the image.AnswerBecause the lens is a diverging lens, its focal length is negativeBecause the lens is diverging, we expect it to form an upright, reduced, virtual image for any object positionUsing the thin lens equation 1/p + 1/q = 1/f <strong>and</strong> solving for q we find thatq =( 1q − 1 ) −1 ()1=p −10.0 cm − 1 −130.0 cm= −7.50 cmThe magnification of the image isM = − q p= −−7.50cm30.0 cm = +0.250This result confirms that the image is virtual, smaller than the object, <strong>and</strong> upright<strong>Light</strong> <strong>and</strong> <strong>Image</strong> <strong>Formation</strong>
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