Homework Help: Lenses Question -- image is to be the same size as the object

1. Mar 18, 2015

Reptic

1. The problem statement, all variables and given/known data
At what distance from an object must a 55mm camera lens be placed if the image is to be the same size as the object?

2. Relevant equations
1/u + 1/v = 1/f
m = v/u
do/di = ho/hi

3. The attempt at a solution
My thinking is that (m) needs to equal 0 therefore v/u needs to equal 0...and that's where i'm up to
or is it just 55mm

1/55 + 1/v = 1/55
0.01 + i/v = 1/55
1/v = 0.01 - 0.01 = 0
1/0 = 1/55...

Any help would be appreciated

Last edited by a moderator: Mar 18, 2015
2. Mar 18, 2015

BvU

Hello Reptic, welcome to PF !

What are all those variables in your equations ? in particular the last one ? Do you need that one ?

Can you say something about v and u if you know m = 1 ? So what does that give when you use the first equation ?

3. Mar 18, 2015

Reptic

I wasn't sure about the last one, thought it might be relevant as it says that object height and image height is the same as image and object distance...i guess not

If m is 1 then would that mean that u and v need to equal 1 to find the answer...i don't know how to write that into the first
equation

is there a specific equation i can either use or rewrite?

Im no closer to an answer unfortunatly

Last edited: Mar 18, 2015
4. Mar 18, 2015

BvU

object and image distance... hadn't you chosen different variable names for those in the first equation ? That way you risk confusion !

And what do you know about ho/hi ?

No, u and v do not have to be equal to one. They only have to be equal to each other, so that v/u = 1. In other "words": v = u.

That means that in the first equation you can replace one of the two by the other, e.g. replace u by v. What do you get when you do that ?

5. Mar 18, 2015

Reptic

Im afraid i am now completely lost:

Very little, in fact nothing

I don't see how interchanging the values of v or u makes a change if they're the same. Then even if i do i don't know how to put them into an equation that gets me towards the correct result.

6. Mar 18, 2015

BvU

If the image is the same size as the object, and you divide the image height by the object height, what do you expect you will get ?

No, interchanging is not very useful, I agree. But it is very useful if you replace one by the other and write down what you get. Why don't you just do it, e.g. replace u by v. What do you get when you do that ?

PS. Do you by any chance have an old reflex camera with a detachable 55 mm lens lying around ? (In fact any lens with a known focal length will do). Place the lens in front of you, a candle on one side and with a piece of paper find where the image is on the other side. Play around until flame and image of flame are about the same size, and at that configuration measure v and u !​

Last edited: Mar 18, 2015
7. Mar 19, 2015

Reptic

1/f = 1/v + 1/u

1/55 = 0.5 x 1/55 + 0.5 x 1/55

0.0181 = 0.009 + 0.009

0.009 = Approx 111 mm??

8. Mar 19, 2015

BvU

What you do is not what we normally call solving.

Solving means substituting variables by expressions from the equations and juggling the symbols until you end up with a usable expression (in symbols) for whatever you want to find.
Then you check the dimension(s). Only then you substitute numbers.
Next: estimate order of magnitude (and approximate value, if possible) and finally you employ a calculator.​

In this excercise the dimension check is almost unnecessary and the calculator isn't needed.

If you replace u by v in
1/f = 1/v + 1/u​
what do you get ?

9. Mar 19, 2015

Reptic

2/v

10. Mar 19, 2015

BvU

Almost. You get 1/f = 1/v + 1/v = 2/v. Now how can you rewrite 1/f = 2/v in such a way that you get something like v = ... in terms of f ?

11. Mar 19, 2015

Reptic

v = 2 x 1/55?

12. Mar 19, 2015

BvU

No way:
1. it now has the wrong dimensions: length on the left and 1/length on the right
2. there is no f in sight. Still too early to put in numbers!
If you have 1/f = 2/v you multiply left and right by f (this is allowed, because f is not equal to zero). That gives you 1 = 2f/v . (Still ok dimensionally !)
Repeat the trick: multiply left and right by v (also allowed: v = 0 can be readily excluded) and tataaa: an exact expression for v ! No need for "approx" and no need for a calculator (I hope )

I have somewhat violated the PF rules, but I'll take the punishment in stride if this helps you understand how to deal with this kind of exercises in a physicist's way

13. Mar 19, 2015

Reptic

hahaha put it like that and nothing seems easier so the answer is just 2f or 110

so to express it in the answer i would say something like

1/f = 1/v + 1/u
∴ 1/f = 1/v + 1/v = 2/v
1/f = 2/v
1 = 2f/v
v = 2f

14. Mar 19, 2015

BvU

Well done ! And it's a good exercise to draw a diagram showing the rays in this special configuration (m = 1)

They show it about 2/3 of the way down in this link

15. Mar 19, 2015

Reptic

Thank you so much for that. I look forward to using that in the future...