Find the position of the lens between two holes and screen

[moenste]

Homework Statement

An opaque card is pierced by two small holes 4 mm apart and strongly illuminated from one side. A lens on the other side of the card focuses images of the holes, 16 mm apart, on a screen 125 cm from the card. Find: (a) the position of the lens, (bc) the focal length. Why are measurements of the object and image sizes in this experiment arrangement unlikely to yield a reliable value for the focal length of the lens?

Answers: (a) 25 cm from card, (b) 20 cm.

2. The attempt at a solution
(b) 1 / u + 1 / v = 1 / f
1 / 25 + 1 / (125-100) = 1 / f
f = 20 cm

Besides that I don't quite understand how to find the 25 cm from the card. I also don't quite understand what do the 16 mm represent.

This is how see it:

We have two holes which are 4 mm apart. We need to find the distance to the lens (a). 16 mm (wrongly on the image cm) is the distance between the projected light.

Having all this I don't know what to proceed with.

For the "Why are measurements of the object and image sizes in this experiment arrangement unlikely to yield a reliable value for the focal length of the lens?" maybe this is because of the two holes and not one?

 

[BvU]

The 16 mm is the distance on the screen between the images of the holes. So you have an enlargement factor. Comes in handy for the v and u !

You verified (b) using the book answer of 25 cm for (a) ?

 

[moenste]

The 16 mm is the distance on the screen between the images of the holes. So you have an enlargement factor. Comes in handy for the v and u !

Re-made the graph:

If it's correct I still don't see what to start with :\

You verified (b) using the book answer of 25 cm for (a) ?

Yes, just checked whether my assumption for (b) was correct.

 

[BvU]

So far we have $\displaystyle {1\over v} +{1\over u}={1\over f} \$ and $u + v = 1.25$ m, so we need one more equation. Did you notice my hint about the enlargement factor ? can you use it to bring in something about one or more of the unknowns ?

And then we are going to come up with a decent drawing. This one looks like a shower design and it's below your standards

 

[moenste]

So far we have $\displaystyle {1\over v} +{1\over u}={1\over f} \$ and $u + v = 1.25$ m, so we need one more equation. Did you notice my hint about the enlargement factor ? can you use it to bring in something about one or more of the unknowns ?

Now I get it. You meant magnification by the "enlargement factor"?

m = h_image / h_objective = v / u
m = (0.016) / (4 * 10^-3) = 4
4 u = v

And because we have u + v = 1.25
u + 4 u = 1.25
5 u = 1.25
u = 0.25 m or 25 cm

Right?

And (b) using the 1 / u + 1 / v = 1 / f as I used in my first post.

And then we are going to come up with a decent drawing. This one looks like a shower design and it's below your standards

This is how see it :).

I get the (a) and (b) parts. But why "Why are measurements of the object and image sizes in this experiment arrangement unlikely to yield a reliable value for the focal length of the lens?"? Is it because there are two holes and so the reliability decreases (one hole would be more reliable)? Also don't quite understand what to start with here.

 

[BvU]

Nice of you to like my compliment (#4).
Reason I bring in the drawing is simple: There are three things you know when drawing object, positive thin lens and image:

1. rays going through focal point come out parallel to the optical axis
2. rays coming in parallel go through focal point
3. rays going through center go through unaffected

Now look at this rule 3 ! That alone is enough to solve this whole exercise (a) "graphically" ! And either one of the others gives the solution for (b)

However, you misinterpreted these rules (I try to keep it as positive as possible).
The lens is convex, so it should bend rays towards the optical axis, not away from it.
Start with rule 3 when re-drawing !
(Hint: is the image upside up or upside down ?)

 

[moenste]

Nice of you to like my compliment (#4).
Reason I bring in the drawing is simple: There are three things you know when drawing object, positive thin lens and image:

1. rays going through focal point come out parallel to the optical axis
2. rays coming in parallel go through focal point
3. rays going through center go through unaffected

Now look at this rule 3 ! That alone is enough to solve this whole exercise (a) "graphically" ! And either one of the others gives the solution for (b)

However, you misinterpreted these rules (I try to keep it as positive as possible).
The lens is convex, so it should bend rays towards the optical axis, not away from it.
Start with rule 3 when re-drawing !
(Hint: is the image upside up or upside down ?)

So the image is upside down like here

 

[BvU]

Nice picture ! I see the 20 cm focal distance to the right of the lens (on the image side).

But not on the object side. You omitted the ray that would follow rule 1 -- to save space ? Or was the 'object' of 23-2 distracting you ?

 

[moenste]

Nice picture ! I see the 20 cm focal distance to the right of the lens (on the image side).

But not on the object side. You omitted the ray that would follow rule 1 -- to save space ? Or was the 'object' of 23-2 distracting you ?

Very sorry for not replying earlier. Continued to study other topics and decided to come back when I'm done.

I think now everything should be correct :). The middle rays go staight to the images, the parallel go through the F and these which go through F' are parallel.

But I still don't quite understand "Why are measurements of the object and image sizes in this experiment arrangement unlikely to yield a reliable value for the focal length of the lens?"

 

[BvU]

I'm proud of your drawing ! Well done. Rapid progress in quality over the various posts ! (*)

Re reliability: Not really sure. Measuring an object size of 4 mm maybe isn't all that easy. Or the image won't be as sharp as desirable ?
Why don't you do the experiment with the lens of your camera or some other lens ?

(*) And here I should probably concede that the post #7 picture was correct as well -- but that's what you get with such high expectations

 

[moenste]

I'm proud of your drawing ! Well done. Rapid progress in quality over the various posts ! (*)

Re reliability: Not really sure. Measuring an object size of 4 mm maybe isn't all that easy. Or the image won't be as sharp as desirable ?
Why don't you do the experiment with the lens of your camera or some other lens ?

(*) And here I should probably concede that the post #7 picture was correct as well -- but that's what you get with such high expectations

:) thank you! And again sorry for a delay in reply. Usually I reply asap.