# Question on Optics

1. Dec 31, 2008

### questions_uk

Hi. "A biconvеx lеns has a focal length of 100mm. an object is placеd at distancеs of 150 mm and 75 mm from teh lens. Sketch a ray diagram to find the position of thе respective images. confirm these positiones via calculatinons.

This is the diagram I have drawn:
http://www.freewebs.com/questionsuk/diagram.JPG [Broken]

Does this seem ok?

However, what equations would be used to confirm the position?

Any help is appreciated.

Last edited by a moderator: May 3, 2017
2. Dec 31, 2008

### dlgoff

Last edited by a moderator: Apr 24, 2017
3. Dec 31, 2008

### questions_uk

Thanks for your help, although how does that equation consider the position of the object from the lens?

4. Dec 31, 2008

### dlgoff

Last edited by a moderator: Apr 24, 2017
5. Jan 1, 2009

### questions_uk

Test

Last edited: Jan 2, 2009
6. Jan 1, 2009

### questions_uk

Test

Last edited: Jan 2, 2009
7. Jan 1, 2009

### Staff: Mentor

Not really. To learn how to draw a ray diagram to calculate the image distance, read this: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/raydiag.html#c1".

Think of the thin lens equation as 1/f = 1/do+ 1/di. Here do is the object distance, the distance between the object and the lens, and di is the image distance, the distance between the image and the lens. Use the equation to solve for di.

You are supposed to figure out the location of the image for two cases: one where do = 75 mm and one where do = 150 mm. (The big difference is that in one case the object is inside the focal length, while in the other it is outside that length.)

Last edited by a moderator: May 3, 2017
8. Jan 1, 2009

### questions_uk

Just a question about the diagram, does this seem more correct?

http://www.freewebs.com/questionsuk/diagram.JPG [Broken] (you might need to refresh)

Test

Last edited by a moderator: May 3, 2017
9. Jan 1, 2009

### Staff: Mentor

Yes, much better. But to get a numerical result you must draw it carefully and to scale.

What does the lens equation tell you?

Last edited by a moderator: May 3, 2017
10. Jan 1, 2009

### questions_uk

Thanks for the reply. For the 150 mm object I worked the image out to be 250 mm from the lens.

For the 75 mm however I've done

1 / 100 = 1 / 75 + 1 / v

= 0.003

= 333 mm (I think it's a negative as it is behind the focal length)

11. Jan 1, 2009

### Staff: Mentor

Redo that calculation.

Redo that more carefully. Your equation is correct, just solve for the unknown.

12. Jan 1, 2009

### questions_uk

How comes? I did 1 / f = 1 / u + 1 / v

1 / 100 = 1 / 150 + 1 / v2

0.01 = 0.006 + 1 / v2

0.01 - 0.006 = 0.004 which is equivalent to 1 / v 2 so

v2 = 0.004

= 250 mm ?

v2 by the way is not v x 2

Thanks.

13. Jan 1, 2009

### Staff: Mentor

Don't round off. That should be 1/150 = 0.006666...

Even better, do this:

1/100 - 1/150 = 3/300 - 2/300 = 1/300 = 1/v2.

14. Jan 1, 2009

### questions_uk

Thanks! For both I got 300 mm although the object at 75 mm is a virtual image.

15. Jan 1, 2009

### Staff: Mentor

Good!

16. Jan 1, 2009

### questions_uk

Also, it has given the index as 1.6 and has asked to calculate the radii of the curvature of the lens.

I've used this equation:

Radii of curvature R = 1 / u + 1 / v = 1 / f (n1 – 1) (1 / R1 – 1 / R2)

Now, I assume it's asking for one radius so I've not included the 1 / R2 in that equation.

To make it simpler the equation is: 1 / f (n1 – 1) (1 / R1 – 1 / R2)

I did: 0.01 = 0.6 [1 / R1] although there may also need to be a -1 / R2 there as well.

I then made R1 to be 6mm.

Any replies are appreciated!

17. Jan 1, 2009

### Staff: Mentor

A more reasonable assumption would be to assume that the biconvex lens is symmetrical--both sides have the same radius. (And be more careful with your arithmetic.)

18. Jan 1, 2009

### questions_uk

Ah yes thanks very much!

One last thing, it says "calculate the longitudinal and transverse magnification of the image if the object is 10 mm high and 5 mm long".

For the longitudinal it has given fo x fi / xo^2

fo is the focal length of the object
fi is the focal length of the image
xo is the distance from the object to the focal length

For the 150 mm object I got -4 and the 75 mm object 16

However, where does the objects dimensions (10 mm high and 5 mm long) come into this? Perhaps it just comes into the transverse equation which is

fo / xo = xi / fi = -yi / yo = xo xi = fo fi

??

19. Jan 1, 2009

### questions_uk

Actually, the yo and yi represent the lengths of the object and its respective image. Not sure about the 5mm long though.

20. Jan 1, 2009

### Staff: Mentor

Only the longitudinal magnification will depend on the object size.