Thin Lens Problem: Solving for Object Distance w/ Magnification

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To determine the object distance for a 50.0-mm-focal-length lens with a 2.00X magnification, the magnification equation (M = -di/do) is used alongside the thin lens equation (1/f = 1/do + 1/di). For a real image, the relationship di = 2(do) allows for the calculation of do. In contrast, for a virtual image, di is negative, leading to the equation di = -2(do). The discussion emphasizes the importance of ray tracing for visualizing the problem and confirming the mathematical solutions.
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Homework Statement



(a) How far from a 50.0-mm-focal-length lens must an object be placed it its image is to be magnified 2.00X and be real?

(b) What if the image is to be virtual and magnified 2.00X?

Homework Equations



Thins Lens Equation: 1/f = 1/do + 1/di

Magnification Equation: M = hi/ho = -di/do

The Attempt at a Solution



I see that we're only given three pieces of information:
  • f=50.0mm
  • M=2.00
  • The image is real

However, I don't know how to incorporate the magnification equation with the Thins Lens Equation.
 
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LastXdeth said:

Homework Statement



(a) How far from a 50.0-mm-focal-length lens must an object be placed it its image is to be magnified 2.00X and be real?

(b) What if the image is to be virtual and magnified 2.00X?

Homework Equations



Thins Lens Equation: 1/f = 1/do + 1/di

Magnification Equation: M = hi/ho = -di/do

The Attempt at a Solution



I see that we're only given three pieces of information:
  • f=50.0mm
  • M=2.00
  • The image is real

However, I don't know how to incorporate the magnification equation with the Thins Lens Equation.

Draw a ray trace - the answer is then obvious, and comes VERY quickley.
 
LastXdeth said:

Homework Statement



(a) How far from a 50.0-mm-focal-length lens must an object be placed it its image is to be magnified 2.00X and be real?

(b) What if the image is to be virtual and magnified 2.00X?

Homework Equations



Thins Lens Equation: 1/f = 1/do + 1/di

Magnification Equation: M = hi/ho = -di/do

The Attempt at a Solution



I see that we're only given three pieces of information:
  • f=50.0mm
  • M=2.00
  • The image is real

However, I don't know how to incorporate the magnification equation with the Thins Lens Equation.

Have you drawn the ray trace yet?
 
Thanks!

I have answered my own question mathematically.

Since the magnification is 2.00, di = (2)(do). I just plugged it in the lens equation to find do.
 
LastXdeth said:
Thanks!

I have answered my own question mathematically.

Since the magnification is 2.00, di = (2)(do). I just plugged it in the lens equation to find do.

Did you get both the real image and virtual image answers?
 
Yes, a virtual image would have a di = -2(do) because the virtual image is upright and in front of the lens.
 
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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