# Ch34Word - Ch. 34 p. 1 CHAPTER 34 Lenses and Optical...

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Ch. 34 p. 1 CHAPTER 34 – Lenses and Optical Instruments 1. ( a ) From the ray diagram, the object distance is about 3 & focal lengths, or 250 mm . O I F F ( b ) We find the object distance from ( 1 / d o ) + (1/ d i ) = 1/ f ; ( 1 / d o ) + (1/88.0 mm) = 1/65.0 mm, which gives d o = 249 mm = 24.9 cm . 2. To form a real image from a real object requires a converging lens . We find the focal length of the lens from ( 1 / d o ) + (1/ d i ) = 1/ f ; (1/285 cm) + (1/48.3 cm) = 1/ f , which gives f = + 41.3 cm . Because d i > 0, the image is real . 3. ( a ) The power of the lens is P = 1/ f = 1/0.275 m = 3.64 D, converging . ( b ) We find the focal length of the lens from P = 1/ f ; – 6.25 D = 1/ f , which gives f = – 0.160 m = – 16.0 cm, diverging . 4. ( a ) We locate the image from ( 1 / d o ) + (1/ d i ) = 1/ f ; (1/18 cm) + (1/ d i ) = 1/24 cm, which gives d i = – 72 cm. The negative sign means the image is 72 cm behind the lens (virtual). ( b ) We find the magnification from m = – d i / d o = – (– 72 cm)/(18 cm) = + 4.0 . 5. ( a ) Because the Sun is very far away, the image will be at the focal point. We find the size of the image from m = h i / h o = – d i / d o ; h i /(1.4 × 10 6 km) = – (28 mm)/(1.5 × 10 8 km), which gives h i = – 0.26 mm . ( b ) For a 50 mm lens, we have h i /(1.4 × 10 6 km) = – (50 mm)/(1.5 × 10 8 km), which gives h i = – 0.47 mm . ( c ) For a 200 mm lens, we have h i /(1.4 × 10 6 km) = – (200 mm)/(1.5 × 10 8 km), which gives h i = – 1.9 mm . 6. We find the object distance from the required magnification (which is negative for a real object and a real image): m = h i / h o = – d i / d o ; ( 2 . 7 0 × 10 3 mm)/(36 mm) = – (8.00 m)/ d o , which gives d o = 0.107 m. We find the focal length of the lens from ( 1 / d o ) + (1/ d i ) = 1/ f ;

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Ch. 34 p. 2 (1/0.107 m) + (1/8.00 m) = 1/ f , which gives f = 0.105 m = + 10.5 cm . 7. ( a ) We find the image distance from ( 1 / d o ) + (1/ d i ) = 1/ f ; (1/10.0 × 10 3 mm) + (1/ d i ) = 1/80 mm, which gives d i = 81 mm . ( b ) For an object distance of 3.0 m, we have ( 1 / 3 . 0 × 10 3 mm) + (1/ d i ) = 1/80 mm, which gives d i = 82 mm . ( c ) For an object distance of 1.0 m, we have ( 1 / 1 . 0 × 10 3 mm) + (1/ d i ) = 1/80 mm, which gives d i = 87 mm . ( d ) We find the smallest object distance from ( 1 / d omin ) + (1/120 mm) = 1/80 mm, which gives d omin = 240 mm = 24 cm . 8. We find the image distance from ( 1 / d o ) + (1/ d i ) = 1/ f ; (1/0.125 m) + (1/ d i ) = – 6.0 D, which gives d i = – 0.071 m = – 7.1 cm (virtual image behind the lens). We find the height of the image from m = h i / h o = – d i / d o ; h i /(1.0 mm) = – (– 7.1 cm)/(12.5 cm), which gives h i = 0.57 mm (upright). 9. ( a ) We see that the image is behind the lens, so it is virtual . ( b ) From the ray diagram we see that we need a converging lens . ( c ) We find the image distance from the magnification: m = – d i / d o ; + 2.5 = – d i /(8.0 cm), which gives d i = – 20 cm. We find power of the lens from ( 1 / d o ) + (1/ d i ) = 1/ f = P , when f is in meters; (1/0.080 m) + [1/(– 0.20 m)] = P = 7.5 D . 10. We can relate the image and object distance from the magnification: m = – d i / d o , or d o = – d i / m . We use this in the lens equation: ( 1 / d o ) + (1/ d i ) = 1/ f ; ( m / d i ) + (1/ d i ) = 1/ f , which gives d i = (1 – m ) f . ( a ) If the image is real, d i > 0. With f > 0, we see that m < 1; thus m = – 2.00. The image distance is d i = [1 – (– 2.00)](50.0 mm) = 150 mm.
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## This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.

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Ch34Word - Ch. 34 p. 1 CHAPTER 34 Lenses and Optical...

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