Focal Length & Position of an object.

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SUMMARY

The discussion focuses on calculating the focal length and position of a lens required to create a half-size image of a 2.0-cm-diameter spider located 3.0 m from a wall. Using the thin-lens equation, the focal length was determined to be 0.261 cm. The position of the lens, measured from the wall, was calculated to be 0.1305 cm. Key corrections were made regarding the use of object distance and image size in the calculations.

PREREQUISITES
  • Understanding of the thin-lens equation (1/f = 1/s + 1/s')
  • Knowledge of magnification equations in optics
  • Familiarity with the concept of object distance (s) and image distance (s')
  • Basic principles of lens optics and image formation
NEXT STEPS
  • Study the derivation and application of the thin-lens equation
  • Learn about magnification and its relationship to object and image sizes
  • Explore different types of lenses and their properties
  • Investigate practical applications of lens calculations in photography and optics
USEFUL FOR

Students in physics, particularly those studying optics, as well as educators and anyone interested in understanding lens behavior and image formation.

frederickcan
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Homework Statement



A 2.0-cm-diameter spider is 3.0 m from a wall.

(a) Determine the focal length of a lens that will make a half-size image of the spider on the wall.

(b) Determine the position (measured from the wall) of a lens that will make a half-size image of the spider on the wall.

Homework Equations



thin-lens equation
1/f= 1/s + 1/s'

The Attempt at a Solution



(a) Used s and s' in the thin-lens equation to find the focal length:
1/2.0cm + 1/0.3cm = 3.83
f= 1/3.83 = .261cm

(b) .261/2 = .1305
 
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For part (a) (Worry about part (b) later)

An additional equation you need is the magnification. It will allow you to use the fact that the image is half the size of the object. What is that equation? Also, you use 2.0 cm for s. You can't do this. Quantity s is the the distance of the spider from the lens. Quantity 2.0 cm is the size of the spider's of the image on the wall.

Also, what is the 0.3 cm in your attempt represent? If it is a mistakenly substituted 3.0 m, then it should not be there. The 3.0 m given in the problem is the distance from the object to its image, i.e. s + s'.

Can you put it together?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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