Geometric optics: Thin lense equation

In summary, the problem asks how many well-focused images of a 2.0-cm-tall candle flame can be formed on a wall 2.0 m away using a lens with a focal length of 32 cm. The relevant equation is 1/f = 1/s' + 1/s, and the lens is converging. The student attempted to use the equation s'+s=2 and s's=2/f, but the latter equation is incorrect as it has units of m^2 on the left side and is dimensionless on the right side. The student should check their derivation of the equation and correct any mistakes.
  • #1
Beth N
41
4

Homework Statement


A 2.0-cm-tall candle flame is 2.0 m from a wall. You happen to have a lens with a focal length of 32 cm. How many places can you put the lens to form a well-focused image of the candle flame on the wall? For each location, what are the height and orientation of the image?

Homework Equations


## \frac {1} {f} = \frac {1} {s'} + \frac {1} {s} ## Where s' is image distance, s is object distance, and f is focal length of the lens. The lens is converging because its focal length is positive.

The Attempt at a Solution


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I just have a hard time approach in the question. I came up with a quadratic formula but still got wrong answer. Apologize for not writing this in Latex I am in quite a rush.

s'+s=2
s's=2/f=2/0.32=6.25

The answer key provided is s′=160 cm, h′=0.50 cm; s′=40 cm, h′=8.0 cm

Thank you so much!
 
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  • #2
Beth N said:
s's=2/f
This equation is incorrect. Check your derivation of it. (Note that the left side of your equation has units of m2 while the right side is dimensionless since you have 2m divided by the focal length in meters.)
 
  • #3
TSny said:
This equation is incorrect. Check your derivation of it. (Note that the left side of your equation has units of m2 while the right side is dimensionless since you have 2m divided by the focal length in meters.)

Is this the correct approach though? I'm not sure what other equation to use
 
  • #4
Yes, your approach is correct. Can you write out the steps that led you to s's=2/f? Then we can identify the specific mistake that you must have made.
 

Related to Geometric optics: Thin lense equation

1. What is the thin lens equation and how is it used in geometric optics?

The thin lens equation is a mathematical formula that relates the focal length, object distance, and image distance of a thin lens. It is commonly used in geometric optics to calculate the position and size of an image formed by a lens.

2. How is the thin lens equation derived?

The thin lens equation is derived from the laws of refraction and the geometry of a thin lens. It takes into account the curvature of the lens and the distance between the object and the lens.

3. Can the thin lens equation be applied to all types of lenses?

Yes, the thin lens equation can be applied to all types of lenses, including convex, concave, and plano-convex lenses. It is a fundamental equation in geometric optics and is used to analyze the behavior of light passing through any type of lens.

4. What is the significance of the thin lens equation in real-life applications?

The thin lens equation is used in a variety of real-life applications, such as in the design of eyeglasses, cameras, telescopes, and microscopes. It allows engineers and scientists to accurately predict the behavior of light passing through lenses and to create precise optical systems.

5. Are there any limitations to the thin lens equation?

The thin lens equation is a simplified model that assumes a thin lens and paraxial rays (rays close to the optical axis). It does not take into account factors such as lens thickness, aberrations, and diffraction. Therefore, it may not be accurate in certain situations, but it is still a useful tool for understanding the basic principles of geometric optics.

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