# Homework Help: More Optics!

1. Feb 17, 2007

### Brewer

1. The problem statement, all variables and given/known data
An object is placed 6 metres from a convex lens of focal length 30cm. A concave lens of focal length 5cm is then placed 20 cm from the concave lens, on the side distant from the object.

Determine the position, magnification and nature of the final image formed (to solve the problem apply the lens-makers formula twice).

2. Relevant equations
Lens makers equation. I assume of the thin lens form (1/p + 1/q = 1/f)

3. The attempt at a solution
Applying the thin lens formula upon the initial lens, it tells me that an image is produced 31.5cm to the right (assuming working left to right) of the lens. That bit I understand.

However, the other lens is placed 20cm to the right of the first, and as a result the image produced by the first lens hasn't formed at this point. So, from what I can understand from my textbook and notes the rays then head back to being paralell.

But, I also saw that a focal length for a diverging lens is negative, so does that mean that I can say that the object is 20cm away from the lens, as the rays start to bend there? Combining this with the focal length of -5cm, and putting into the thin lens equation will give an image distance of -0.04m, or 4 cm to the left of the concave lens.

So does this mean that the image will be formed inbetween the two different lenses?

If this is the correct distance, does this mean that the magnification is 1/95th of the original size? And the image will also be inverted (I assume that this is what the question means by the nature of the image).

Is this right, or have I gone wrong somewhere?

2. Feb 17, 2007

### Staff: Mentor

Here's a tip for you. When working with mutliple lenses, the image from the first lens becomes the object for the second. Note that in this case, the "object" for the second lens is to the right of the lens. So what does that tell you about the sign of the object distance?

3. Feb 17, 2007

### Brewer

Does that mean that the sign becomes negative?

But i think what you said first was my problem. Because the image from the first lens is formed to the right of the second lens, doesn't that mean that the rays that form image1 have to pass through the second lens, and as a result doesn't that refract those rays so that the image is formed in a different place than I originally calculated?

But apart from the sign convention am I headed along the correct lines do you think?

4. Feb 17, 2007

### Staff: Mentor

Yes, in applying the thin lens formula to the second lens, the object distance will be negative.

I don't understand how you calculated your final image position. (But maybe I just need more coffee.)

Yes, the light that would have produced an image from the first lens is prevented from doing so by the second lens. But applying the thin lens formula twice incorporates all of that. (Recall that the thin lens equation is derived by considering in detail the refractions at the surfaces of the lens.)

I would do it in two steps, first one lens (done!) and then the second lens (do it right).

5. Feb 17, 2007

### Brewer

So,

I've calculated the first lens? And for the second lens I can ignore the fact that the lens is in the way when the image is formed?

6. Feb 17, 2007

### Staff: Mentor

Yes. Giving the image distance is a shorthand way of describing the action of the first lens. Use that as input for the second lens (in the form of an object distance) and then you can calculate the effect of the second lens, giving the final image location.

7. Feb 17, 2007

### Brewer

Ok then, so using a negative focal length for the second length (is this correct - I think its right from what I read in my text book, but I'm not 100% sure) I calculate the image to be at -0.088m from the second lens. That means that the image is formed between the two lenses isn't it? My text book says:

When the image is on the same side of the reflecting or refracting surface as the outgoing light, the image distance is positive, otherwise it is negative.

Now I would assume that the sign conventions set up for the first lens will then hold true throughout, so that they are based upon the directions of the incoming and outcoming light of the first lens, which then leads me to this conclusion that the image is formed in between the two lenses, about 9cm from the second lens.

8. Feb 17, 2007

### Staff: Mentor

Right. It's a concave (diverging) lens so it has a negative focal length.
Right.
This is true--standard sign convention.

Not sure what you mean by this. You are using the same sign convention throughout.
Yes, which you already stated above. The final image is 8.8 cm to the left of the 2nd lens, which puts it between the two lenses.

What kind of image must it be?

9. Feb 17, 2007

### Brewer

It's a virtual image isn't it? Because it it formed by a diverging mirror, and is within the focal length of the converging mirror.

Is this what is meant by the nature of the mirror?

I assume to find the magnification of the image, I find the magnification due to lens 1 and then multiply by the magnification of the second lens.

Thank you for your help. Things are starting to seem much clearer now.

10. Feb 17, 2007

### Staff: Mentor

It's a virtual image because it's on the left of the final lens, whereas the actual light is on the right.

You mean nature of the image. I'd say yes.

Yes.

Good!