Ray-Tracing vs. Depending on Formulas

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To achieve a sharp image of a lightbulb on a screen using a lens, the lens must be moved based on the relative positions of the object and the screen. The formula 1/s + 1/s' = 1/f indicates that to decrease the distance d between the object and image, adjustments to s (the distance from the lens to the object) are necessary. If the lens is closer to the object than the screen, moving it away from the object will help focus the image. Conversely, if the lens is closer to the screen, it should be moved away from the screen. The discussion highlights the ambiguity in the problem due to unspecified distances, which affects the clarity of the solution.
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A lens forms a real image of a lightbulb, but the image of the bulb on a viewing screen is blurry because the screen is slightly in front of the image plane. To focus the image, should you move the lens toward the bulb or away from the bulb.

Formulas:
1/s+1/s'=1/f

Well, in the book, the answer is move away. It says we need to decrease s', so increase s.

But I made drawings using ray-diagrams, and what I get is that, for an object further away from the focal point of a converging lens, moving the lens closer to the object would make more sense... The image plane actually gets closer to the object.

There seems to be a contradiction here...
 
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According to the problem text, the object and the screen are at fixed positions, and the lens has to be moved to give sharp image. The distance between the object and image is d=s+s', and the distance between the screen and object is D. It is said that the screen is slightly in front of the image plane, which means D<d. You have to decrease d=s+s'. Derive the expression for d=s+s' and see if you need to increase or decrease s so as d decrease.

ehild
 
I mean to see a little snag here: usually there are two positions of the lens that yield a sharp image (provided d > 4f): s and s' are interchangeable.

If the lens is closer to the object than to the screen, you can reduce s+s' by moving the lens away from the object.

If the lens is closer to the screen than to the object, you can reduce s+s' by moving the lens away from the screen.

Since we haven't been told which s is the distance between screen and lens and which is the distance between object and lens, nor whether the distance between screen and lens is smaller than the distance between object and lens, I end up considering the question ambiguous. Unless there's more information in the exercise than lingy mentions.
 
@BvU: yes, it depends on the position of the object what we should do with the lens in order to get a sharp image. But I hoped that the OP would find it out.

ehild
 
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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