# Optical Instruments I - Thin Lens

1. Jun 30, 2009

### OsDaJu

1. The problem statement, all variables and given/known data

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 [Broken]

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.)

2. Relevant equations
1\o + 1\i = 1\f

3. 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?

Last edited by a moderator: May 4, 2017
2. Jun 30, 2009

### alphysicist

Hi OsDaJu,

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?

Last edited by a moderator: May 4, 2017
3. Jun 30, 2009

### OsDaJu

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!

4. Jun 30, 2009

### alphysicist

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

No, the image distance is not zero. Remember the the equation uses reciprocals. So what is i, and where then is the image?

Last edited: Jun 30, 2009
5. Jun 30, 2009

### jualin

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.

6. Jun 30, 2009

### alphysicist

Hi jualin,

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?

7. Jun 30, 2009

### OsDaJu

Well if I take the limit as i goes to infinity then it equals zero. So the image is at infinity?

8. Jun 30, 2009

### alphysicist

That sounds right to me.

9. Jun 30, 2009

### jualin

Thank you!

10. Jun 30, 2009

### OsDaJu

Thanks.

11. Jun 30, 2009

### alphysicist

Sure, glad to help!

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