Optics: What is the speed of the sprinter's image?

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SUMMARY

The discussion centers on calculating the speed of a sprinter's image using a 150 mm focal length lens while the sprinter runs at 5 m/s. The formula used is the lens formula, (1/f) = (1/s') + (1/s), where f is the focal length, s' is the image distance, and s is the object distance. The calculated image distance is approximately 0.152 m, leading to a magnification of -0.0152. Consequently, the speed of the sprinter's image is determined to be 76 mm/s.

PREREQUISITES
  • Understanding of the lens formula in optics
  • Knowledge of magnification concepts
  • Basic principles of relative motion
  • Familiarity with units of measurement in optics (meters, millimeters)
NEXT STEPS
  • Study the lens formula in detail, focusing on its applications in photography
  • Explore the concept of image speed in relation to object motion
  • Learn about different types of lenses and their focal lengths
  • Investigate the effects of distance on image size and speed in optics
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This discussion is beneficial for physics students, photographers interested in optics, and anyone studying motion and image formation in photography.

bcjochim07
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Homework Statement


A sports photographer has a 150 mm focal length lens on his camera. The photographer wants to photograph a sprinter running straight away from him at 5 m/s. What is the speed of the sprinter's image at the instant the sprinter is 10 m in front of the lens?


Homework Equations





The Attempt at a Solution



Here's my best guess:

(1/f)=(1/s')+(1/s) (1/.150m)= (1/s') + (1/10m) s'=.152, so the absolute value of the magnification is (.152/10m)= -.0152

and (.0152)(5m/s)= .076 m/s = 76 mm/s Is this anywhere close to correct?
 
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