What is the distance between a diverging lens and a filament?

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Homework Help Overview

The discussion revolves around a problem involving a diverging lens with a focal length of 5 cm and its relationship to a filament. Participants are attempting to determine the distance between the lens and the filament, while exploring the implications of the lens's properties.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are discussing the definitions of variables in the lens equation, questioning whether the object distance (do) can be equal to the focal length, and exploring the implications of the lens being placed on the filament.

Discussion Status

The discussion is ongoing, with participants clarifying terms and equations. Some have raised questions about the setup of the problem and the assumptions regarding the distances involved, while others are exploring different interpretations of the question.

Contextual Notes

There is ambiguity regarding the placement of the lens relative to the filament, with some participants suggesting that the lens may be in front of the filament rather than directly on it. This has led to confusion about the expected distance measurement.

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Homework Statement


A diverging lens with focus 5 cm is placed on a filament. Find the distance between the lens and the filament

Homework Equations


[tex]\frac{1}{f}=\frac{1}{di}+\frac{1}{d}[/tex]

The Attempt at a Solution


Focus of diverging lens is negative so, f = -5 cm. I think the question is asking about di. What is d ? Can it be infinity because the filament gives light until far enough distance ?

Thanks
 
Last edited:
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From what i recall di = initial distance of object and d is the final. However, to solve your question i won't be as much as of a help as others. Sorry.
 
Hi Maroc

Ah thanks for clarifying my equation. I usually use d as distance of object and d' as distance of the image. I think it's better to use your notation.

Thanks :)
 
No problem songoku,

Just like to help around. If there is no help here try wiki or other info sites regarding this matter.
 
If you real positive sign convention,
the equation becomes
-1/f = 1/do + 1/di, where do is the object distance and di is the image distance.
Here do = f = 5 cm
 
Hi Mr. rl.bhat

Why does do = 5 cm ? Based on your equation, I think the question is asking about do, the distance between the lens and filament.

Thanks
 
songoku said:
Hi Mr. rl.bhat

Why does do = 5 cm ? Based on your equation, I think the question is asking about do, the distance between the lens and filament.

Thanks

he rearranged the equation to get 5cm
 
Hi Maroc

Oh I see. So like I said, di (the distance of image) = infinity, right ?

Thanks
 
songoku said:
Hi Maroc

Oh I see. So like I said, di (the distance of image) = infinity, right ?

Thanks

In the given problem there are two unknowns. By a diverging lens you cannot get an image at infinity. So I thought the object is at the focus of the diverging lens. In the MCQ most of the students will select the image at infinity, which is wrong.
 
  • #10
Hi Mr. rl.bhat

Oh I see. The image of diverging lens should be located between the focus and the center of the lens.
But I don't understand why you thought the object is at the focus. I think it's possible the object (in this case I suppose the object is the filament) is located between f and 2f or further than 2f and the question is asking about this distance.

Thanks
 
  • #11
I'm confused. The question says that the lens is placed on the filament, then asks for the distance between the lens and the filament? If one is on top of the other, wouldn't the distance be 0?
 
  • #12
Hi ideasrule

I'm not sure about that, but that's the question. Or maybe the lens is placed in front of a filament in a lightbulb?

Thanks
 

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