1. Jan 19, 2014

1. The problem statement, all variables and given/known data
I have many questions.Once I get the answer of the first one,I will proceed with the second and so on.
This is really not a home work question.But....
So,In my text book as well as my teacher defines the focal point as :The point where image of a distant(Parallel rays) object is formed.

As there are parallel rays,All the rays will converge to a one single point.So it is not an image.But why does the text book defines like that?

I used a magnifying glass and saw that moon produces a nice image but the sun produce a dot.
But the rays from moon is also considered parallel

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2. Jan 19, 2014

### ehild

The parallel rays meet at the focal point. Both the Moon and the Sun are so far away, that their images forms in the focal point practically.
Still, their image is not a single dot.
Image is formed when the rays emerging from a point of the object meet in a single point again.
The rays arriving at the lens from a point of the Moon (Sun) are not quite parallel, and they meet and form a point of the image.

Assume you have a lens with focal distance of 1 m. The Moon is 385000 km away, and its diameter is about 3500 km. What is the image distance and the size of its image?
The distance of Sun is about 150 million km, and its diameter is about 1.4 million km. Where does its image form and what is the image size?

ehild

Last edited: Jan 19, 2014
3. Jan 19, 2014

Ok.So they are not actually parallel.

Actually I haven't studied that kind of things yet.
So for second question...

4. Jan 19, 2014

As you see,the rays from a point of an object converges to another point to form an image.
But the green and red lines also meet.What will happen there?Will an image be formed?

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5. Jan 19, 2014

### rude man

Put an object of height h and position p=10m to the left of a convex lens of say f = 1m focal length. Draw rays from the top and bottom of the object to the image, which will be to the right of the lens, inverted, at a distance q = pf/(p-f) and of height hq/p = hf/(p-f) = h/(10-1) = 0.111h (all quantities > 0).

Now double the size of the object and double p. Draw rays from the top and bottom of the object again. q will shrink a bit but the image size is almost as big as before: 2h/(20-1) = 0.105h. Notice that the wavefront is closer to being parallel than before.

You can extend this to larger & larger h and p, keeping h/p constant, and the rays will focus closer and closer to the f plane but the height of the image remains almost unchanged. In the limit for p and h → ∞, but keeping h/p = 0.1, the image height is hf/p = 0.100h.

So with for example the Sun, h and p are huge, the wavefront is essentially completely parallel, but the object still focuses to show the correct image. The image forms at the focal plane but is still intact.

Your instructor should have used the term 'focal plane' rather than 'focus'.

6. Jan 19, 2014

### rude man

No.

F is in the focal plane. The red and green rays would converge there if and only if the object distance is infinity.

The red and green rays don't 'see' each other. This is based on the principle of superposition. One ray does not affect another.

EDIT: I should rephrase that: if you place a screen on the image plane where all the green beams meet, all the red beams meet, etc. you will get a sharp image.

If you place the screen at F in your illustration you will also get an image, but since there the green and red beams meet the image will be all blurry.

Last edited: Jan 19, 2014
7. Jan 19, 2014

### ehild

8. Jan 20, 2014

OK.Now I understand
Is there any kind of nice software which simulates what happens to image,rays etc when the object moves away or close to the lens etc??

9. Jan 20, 2014

### Malverin

I found simple simulation here

and Google shows over 2 milion results.
I think you will find what you need there.

10. Jan 20, 2014