Optical Instruments I - Thin Lens

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Homework Help Overview

The discussion revolves around a problem involving a thin lens with a focal length of 8 cm, used in conjunction with a plane mirror and a pinhole for the purpose of auto-collimation. Participants are tasked with determining the placement of the pinhole to achieve this setup and further exploring the concept of focal length through image formation of a distant object, specifically the sun.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the placement of the pinhole at the focal point of the lens and question the implications of this setup. There is confusion regarding the object distance when the object is at the focal point, and participants explore the relationship between object distance, image distance, and focal length using the lens formula. Some participants express uncertainty about the terminology required for the answer.

Discussion Status

The discussion is active, with participants offering insights and questioning each other's reasoning. There is a focus on clarifying the relationship between object distance and image distance, with some guidance provided regarding the use of the lens formula. Multiple interpretations of the image distance are being explored, particularly concerning the implications of the object being at the focal point.

Contextual Notes

Participants are working under the constraints of a homework assignment, which includes specific terminology requirements for answers. There is an ongoing exploration of assumptions related to the setup of the optical system and the definitions of focal length and image distance.

OsDaJu
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Homework Statement



part1: In the Figure below a lens of focal length 8 cm is placed between a plane mirror and a pin hole. The other side of the pin hole is illuminated by a light source (not shown), so that the light passes through the hole and is then collected by the lens. If the pin hole is at the focal point of the lens, then the light that passes through the lens will be parallel. When the parallel rays hit the plane mirror, the reflected rays will also be parallel, and when they pass through the lens again the rays will be focused at the pin hole. This is called the auto-collimation method for finding the focal length of a lens. For this part, the object is to find the location to place the lens and the pin hole to obtain auto-collimation. Question: Where should the pin hole be placed so that it is at the focal point of the lens?
I got part 1

http://img188.imageshack.us/img188/6408/randomlabel.png

Part 2:
The next part of this exercise examines further the concept of focal length.

The lens in the last part is used to form the image of the sun. What is the image distance? (Recall that the focal length is 8 cm.) An object is located at the focal point of the lens. The image of the object is located at [...] (This part is asking for a word, not a number.)

Homework Equations


1\o + 1\i = 1\f

The Attempt at a Solution


I need help with part B.
I don't know where the object is. I assumed it was at the plane mirror therefore the object distance was 4 cm and the focal was 8 cm (given). which was wrong and I'm not sure about the word that they asked for. I tried "focus," "focal length" and "collimation." Why am I wrong?
 
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Hi OsDaJu,

OsDaJu said:
http://img188.imageshack.us/img188/6408/randomlabel.png

Part 2:
The next part of this exercise examines further the concept of focal length.

The lens in the last part is used to form the image of the sun. What is the image distance? (Recall that the focal length is 8 cm.) An object is located at the focal point of the lens. The image of the object is located at [...] (This part is asking for a word, not a number.)

Homework Equations


1\o + 1\i = 1\f


The Attempt at a Solution


I need help with part B.
I don't know where the object is. I assumed it was at the plane mirror therefore the object distance was 4 cm and the focal was 8 cm (given). which was wrong and I'm not sure about the word that they asked for. I tried "focus," "focal length" and "collimation." Why am I wrong?

I'm assuming you're asking about the second part of Part 2; in that case they tell you where the object is --at the focal point of the lens (and I believe the plane mirror has been removed). If the object is at the focal point of the lense, what is the object distance? Then, using your equation, what would the image distance be?
 
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If the object is at the focal point then the object is at the same distance as the focal length, 8 cm. Is this correct? Therefore the image distance would be zero? And what would be the answer for the "word" that they are asking me? Thank you for the quick reply!
 
OsDaJu said:
If the object is at the focal point then the object is at the same distance as the focal length, 8 cm. Is this correct?

Yes, so in your equation both f and o are the same.

Therefore the image distance would be zero? And what would be the answer for the "word" that they are asking me? Thank you for the quick reply!

No, the image distance is not zero. Remember the the equation uses reciprocals. So what is i, and where then is the image?
 
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I have the same question. If you say that "f" and "o" are the same therefore:

1/o+1/i=1/f

or if o=f then

1/f + 1/i = 1/f

so 1/f - 1 /f = 1/i

Wouldn't the "i" be zero? Or would it be the twice the focal length "2f"? Also what would be the answer for the word that is being asked? Thank you for taking your time to reply.
 
Hi jualin,

jualin said:
I have the same question. If you say that "f" and "o" are the same therefore:

1/o+1/i=1/f

or if o=f then

1/f + 1/i = 1/f

so 1/f - 1 /f = 1/i

Wouldn't the "i" be zero? Or would it be the twice the focal length "2f"? Also what would be the answer for the word that is being asked? Thank you for taking your time to reply.

Look at your equation:

1/f - 1/f = 1/i

Go ahead and do the left side:

0 = 1/i

So this is definitely not saying that i=0 (since the equation deals with reciprocals). Where is this equation saying that the image is?
 
Well if I take the limit as i goes to infinity then it equals zero. So the image is at infinity?
 
OsDaJu said:
Well if I take the limit as i goes to infinity then it equals zero. So the image is at infinity?

That sounds right to me.
 
Thank you!
 
  • #10
Thanks.
 
  • #11
Sure, glad to help!
 

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