Diverging lens on filament

  • Thread starter songoku
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  • #1
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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:

Answers and Replies

  • #2
Maroc
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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.
 
  • #3
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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 :)
 
  • #4
Maroc
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No problem songoku,

Just like to help around. If there is no help here try wiki or other info sites regarding this matter.
 
  • #5
rl.bhat
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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
 
  • #6
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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
 
  • #7
Maroc
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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
 
  • #8
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Hi Maroc

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

Thanks
 
  • #9
rl.bhat
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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
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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
ideasrule
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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
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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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