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Phys_219_Lecture_16_

# Phys_219_Lecture_16_ - Lecture 16Lecture 16-1 The Eye q 2.5...

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Lecture 16 Lecture 16-1 The Eye q 2.5 cm (fixed) 1 1 depends on p 11 pq f  f depends on p p => f to keep q at 2.5 cm

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Lecture 16 Lecture 16-2 Compound Lenses 1 q 1 > 0 11 1 p qf  p > 0 =- 0 1 1 p 2 q 1 < 0 12 pq f f 1 qqf 2 q > 0 Power adds up in compound lenses
Lecture 16 Lecture 16-3 Corrective Lenses Hyperopia : A person 11 1 1  1 ype opia :p e s o has near point = 75 cm Can’t focus near.  * 1 41.33 0.75 0.025 E D f  12 pq f f 1 f 1 1 4 (* means most strained eye) Correct to 25 cm by conv. lens Myopia(nearsighted) g far point = 40 cm Can’t focus far. Use a lens of +2.67D   * 44 0.025 0.25 E L D ff  e.g., far point = 40 cm # 1 42.5 0.40 0.025 E D f ( # means least strained eye) Corrected by diverging lens # 1 1 40 0.025 E L D Use a lens of -2.5D

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Lecture 16 Lecture 16-4 Physics 219 – Question 1 – March 9, 2011. John is near-sighted with his far point distance eing 1 m What is the focal length f the being 1 m. What is the focal length f of the corrective lens he should wear. (It should move the far point to .) A. - 1 m . +1m  # 11 1 41 1 0.025 D f  B. 1 m C. - 25 cm 25cm E ( # means least strained eye) D. + 2.5 cm E. - 50 cm # 1 1 40 0.025 E L D ff Use a lens of -1D (f= -1m)
Lecture 16 Lecture 16-5 Magnifying Lens n object is placed near the focal point of a magnifying lens s An object is placed near the focal point of a magnifying lens. Its (virtual) image is infinite distance away and can be seen by the relaxed eye. The angle subtended by the image is tan = y/d 0 y/f . Without

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Phys_219_Lecture_16_ - Lecture 16Lecture 16-1 The Eye q 2.5...

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