Ray-Tracing vs. Depending on Formulas

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Homework Help Overview

The discussion revolves around the optics of a lens forming a real image of a lightbulb, specifically addressing the conditions under which the image can be focused clearly on a viewing screen. The problem involves understanding the relationship between object distance, image distance, and focal length, as described by the lens formula.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of moving the lens closer to or further from the object to achieve a focused image. There is a discussion about the apparent contradiction between the formulaic approach and ray-diagram analysis. Questions arise regarding the fixed positions of the object and screen, and how these affect the distances involved.

Discussion Status

Participants are actively engaging with the problem, examining different interpretations of the lens positioning and the resulting image clarity. Some guidance has been offered regarding the relationships between distances, but ambiguity remains due to insufficient information about the specific distances involved.

Contextual Notes

There is a lack of clarity regarding the exact distances between the lens, object, and screen, which complicates the analysis. The problem's constraints suggest that the object and screen are fixed, but the specific distances are not defined, leading to varying interpretations among participants.

alingy1
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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
 

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