How Do You Manipulate Lenses for Optimal Image Placement?

AI Thread Summary
To obtain an image as far away from the lens as possible, the object should be placed at the focal length, which results in an image at infinity. When an image focused on a slide screen is moved towards the lens, the lens must be moved away to maintain focus. The thin lens equation, 1/do + 1/di = 1/f, is crucial for understanding these relationships. Moving the lens away compensates for the decrease in the image distance (di) while keeping the focal length (f) constant. Understanding these principles is essential for optimal image placement with lenses.
jvdamdshdt
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Homework Statement



1.Where would you place an object to obtain an image as far away from the lens as possible?

2. Suppose you have an image focused on a slide screen. You moved the image towards the lens. Which way would you move the lens to keep the image in focus? Explain

Homework Equations



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The Attempt at a Solution



For 1 I am thinking far away from the image as possible would be at the focal length but do they mean a real image? That would be between the focal length and twice the focal length

For 2 I am thinking you would move the lens away to keep the image in focus.
 
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jvdamdshdt said:

Homework Statement



1.Where would you place an object to obtain an image as far away from the lens as possible?

2. Suppose you have an image focused on a slide screen. You moved the image towards the lens. Which way would you move the lens to keep the image in focus? Explain

Homework Equations



None

The Attempt at a Solution



For 1 I am thinking far away from the image as possible would be at the focal length but do they mean a real image? That would be between the focal length and twice the focal length

For 2 I am thinking you would move the lens away to keep the image in focus.

Are you familiar with the thin lens equation? It can be used to answer both of these questions.

For question 1: imagine you had parallel rays coming in "from infinity." They would be bent by the lens and all converge at the focal point on the other side. Now, reverse time (or equivalently, just reverse the direction of propagation). So, the origin of the rays becomes the destination, and vice versa. So, rays would diverge from the focal point, hit the lens, and be bent to be all parallel on the other side. So where would the image be in this situation? Is it possible to get farther away from the lens than that? :wink:

For question 2: I repeat: do you know the thin lens equation?
 
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I do know the lens equation. 1/do + 1/di = 1/f . In question one the object at the focal length will produce an image at infinity. For question two I don't know if I am remotely close to getting the answer.
 
jvdamdshdt said:
I do know the lens equation. 1/do + 1/di = 1/f . In question one the object at the focal length will produce an image at infinity. For question two I don't know if I am remotely close to getting the answer.

For question 2, you are reducing di and keeping f the same (since the lens curvature doesn't change).
 
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