How far is the lens from the slide if the image is virtual?

1. Jul 20, 2007

floridianfisher

1. The problem statement, all variables and given/known data A transparent photographic slide is placed in front of a converging lens with a focal length of 2.44 cm. The lens forms an image of the slide 13.2 cm from the slide.
How far is the lens from the slide if the image is virtual?
nearer possible distance =

2. Relevant equations
1/p + 1/q = 1/f

3. The attempt at a solution
I was able to find the distance of the lens from the slide if the image was real by 1/x + 1/(13.2-x) +(1/2.44) and solving for x.
However I assume that since the object is virtual the image is between the slide and lens. So, P= 13.2 + x and q=-x while f=2.44. When I plug this into the thin lens equation the solution is wrong
How is The attempt wrong

2. Jul 20, 2007

G01

This isn't the case in this situation. You have a virtual image 13.2 cm away from the object, and it is formed by a converging lens. This means the object is closer to the lens than the focal point, and the virtual image will appear behind it. So, the image distance is a negative quantity, since it is on the same side of the lens as the object.

Try starting the problem this way. Put the object a distance S away from the lens, where S is shorter than the focal length. After you do this, find the image distance in terms of S, and then use the thins lens equation. See how your answer comes out now. Good Luck!

P.S. I do not know what you mean by "nearer possible distance=". If this is another part of the question, can you elaborate please?

3. Jul 20, 2007

floridianfisher

Here is the whole question. I got all of a and the second part of b
(a) How far is the lens from the slide if the image is real? (Enter 'none' in the second box if there is only one answer.)
nearer possible distance = correct check mark cm
further possible distance = correct check mark cm

(b) How far is the lens from the slide if the image is virtual? (Enter 'none' in the second box if there is only one answer.)
nearer possible distance = wrong check mark cm
further possible distance = correct check mark cm

4. Jul 20, 2007

G01

OK, thanks. That helps. Now, when I solved this problem I end up with a quadratic equation at one point. You should end up with two answers for S from this equation.(One may not be physical, remember.) Did you end up in a situation like this?

Last edited: Jul 20, 2007
5. Jul 20, 2007

floridianfisher

Okay so I got this diagram
f----------p-------q----lens
distance from the lens to f =2.44cm
distance from lens to p= 13.2+s
distance from lens to q = s
then solving for s I get s= 2.88 cm or s= -11.2cm
Since q is supposed to be negative for virtual images the answer is 11.2 cm?

6. Jul 20, 2007

G01

I get numbers similar to those, 2.1, and -14. I did round a little, but just to be sure, did you remember to make the whole quantity of "p" negative, since it is one the same side of the lens as the object? EDIT: Forgot we were solving for object distance, not image distance, removed error on my part.

Last edited: Jul 20, 2007
7. Jul 20, 2007

floridianfisher

I used
1/(13.2+x) + 1/x = 1/2.44 for those answers
I just tried it the way I believe your talking about which is
1/-(13.2+x) +1/x = 1/2.44 and I got -2.1 and -15.3
I am confused though

8. Jul 20, 2007

G01

That negative sign has to be there. It tells you that the image is one the same side of the lens as the object. That is what defines a negative image distance in ray optics.

Now those answers are what I got, more or less. Remember that you solved for S, and S has to be less than the focal length, which is 2.44. So this tells us that the answer must be 2.1 . Now you should be able to find the image distance using S.

9. Jul 20, 2007

floridianfisher

I plugged in p= 13.2 +2.1 =15.3 but the answer is wrong?

10. Jul 21, 2007

G01

OK. That may very well be the image distance, but I re-read the question, and they are looking for the object distance for the answer. Sorry if I confused you. I thought the final answer was the image distance not the object distance. Somehow I lost track. Try the object distance of 2.1 as the answer.

11. Jul 21, 2007

floridianfisher

ok thats I got it. Thanks for your help!

12. Jul 21, 2007