# Optics, image

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## Homework Statement

Localize the image of an object situated at 1.2 m from a crystal ball with a diameter of 0.2 m (n=1.5).

## Homework Equations

None given. But Snell's law I believe.

## The Attempt at a Solution

I'm totally stuck at starting. Does the size of the object matters? Should I assume it as 1 or 2 dimensional? As a point? Or it doesn't matter? The image should be on the other side from the ball, right? I mean the image is real and not inside the ball, right?
Thanks for getting me started.

Homework Helper
First step is to calculate the focal length of a sphere (or google it)

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First step is to calculate the focal length of a sphere (or google it)

Ok thanks for the tip. I found http://www.math.ubc.ca/~cass/courses/m309-04a/sol4.pdf. I'm having a hard time understanding what is the transfer matrix (http://en.wikipedia.org/wiki/Ray_transfer_matrix_analysis), specially what are A, B, C and D in the wikipedia article. In my class notes they are partial derivatives...
I've Born's book on Optics, but I didn't find anything on transfer matrices.
So instead of assuming that f=R/2, I must be able to calculate it.

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Homework Helper
focal length of a ball lens is = nD / 4(n-1) measured from the center of the sphere
(don't know if you are expected to derive this as part of the question)

f = R/2 is only for thin lenses, otherwise the focal point of a sphere would be inside the sphere

Gold Member
focal length of a ball lens is = nD / 4(n-1) measured from the center of the sphere
(don't know if you are expected to derive this as part of the question)

f = R/2 is only for thin lenses, otherwise the focal point of a sphere would be inside the sphere

Thanks once again, and yes I have to derive everything. Should I use some transfer matrix?

Homework Helper
Probably easy to derive it from just trig.
then you can just use the normal 1/f, 1/u, 1/v formula

Gold Member
Probably easy to derive it from just trig.
then you can just use the normal 1/f, 1/u, 1/v formula
The problem is that I don't really know how light rays are affected by the sphere.
I post an picture I scanned from my draft, but I don't think it's right. I did the sketch for n=1 for air and n=1.5 for the crystal.
I get it all wrong I guess, it seems the focus is inside the sphere and that the image is smaller than the object and I'd bet that it should have the same dimensions, i.e. magnification of 1.

#### Attachments

• escanear0002.jpg
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Homework Helper
Sorry my post seems to have disappeared - network issues?

Your drawing is for a sphere with a very high refractive index, it's certainly possible to make a spherical lens where the focus is inside - it's useful as a reflector.

A better diagram might be:
http://www.edmundoptics.com/imagelib/techsup/reform/opt028a6.gif [Broken]

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Gold Member
Sorry my post seems to have disappeared - network issues?

Your drawing is for a sphere with a very high refractive index, it's certainly possible to make a spherical lens where the focus is inside - it's useful as a reflector.

A better diagram might be:
http://www.edmundoptics.com/imagelib/techsup/reform/opt028a6.gif [Broken]
[/URL]

Ok thanks.
In fact I followed the instruction to draw such a sketch, i.e. a sphere with an refractive index of 1.5 submerged in vacuum (n=1). But ok, I'll stick with your picture.
I'll come back to this problem later, I'll do some electromagnetism exercises then read an optics book and finally try this exercise again.
Thanks for all and be sure I'll ask my doubts here.

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Gold Member
I'm really lost on this problem!
I just checked out in wikipedia the formula of the lensmaker when the lens has a big thickness. I probably can consider the sphere as being 2 lenses with thickness 0.1 m each...
I just don't know how to solve the problem. Should I draw an arbitrary sketch and figure out everything via some trigonometry? Should I use the matrix method?
I've talked to a friend and he considered the object as a point.
Anyway I repeat but it seems that the focus is inside the sphere. Well, that is if the object has the same proportions compared to the lense as in my sketch. I used very carefully a method to draw how a lense with refractive index of 1.5 would refract the rays from an object.
I'm kind of desperate about this problem which is very simple for all the other people in my class.

Edit: I just tried to use the formula $$\frac{1}{f} = (n-1) \left[ \frac{1}{R_1} - \frac{1}{R_2} + \frac{(n-1)d}{n R_1 R_2} \right]$$ which gave me 0. Meaning that $$f=\pm \infty$$. Impossible of course.
I considered that f=f1+f2 where f1 is the focus of the left half sphere and f2 is the focus of the right half sphere. I separated the sphere into 2 half spheres and considered each half as a lens.

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las3rjock
Born & Wolf is way too advanced a book to be using to solve a problem like this, but the equation you want is in section 4.4.1 (in the sixth edition--it may be different in other editions), "Refracting surface of revolution," and it's the one that looks like this
$$\frac{n_1}{z_1}-\frac{n_0}{z_0} = \frac{n_1 - n_0}{r}$$
where z0 is the displacement of the object from the spherical interface, z1 is the displacement of the image from the spherical interface, and r is the radius of curvature spherical interface (r > 0 if the interface appears convex to light impinging upon it from the -z direction, and r < 0 if the interface appears convex to light impinging upon it from the -z direction). You apply this formula twice: first to locate the image formed by the front surface of the sphere, and then to locate the image formed by the rear surface of the sphere (by considering the image formed by the front surface of the sphere to be the object that is imaged by the rear surface of the sphere).

Gold Member
Born & Wolf is way too advanced a book to be using to solve a problem like this, but the equation you want is in section 4.4.1 (in the sixth edition--it may be different in other editions), "Refracting surface of revolution," and it's the one that looks like this
$$\frac{n_1}{z_1}-\frac{n_0}{z_0} = \frac{n_1 - n_0}{r}$$
where z0 is the displacement of the object from the spherical interface, z1 is the displacement of the image from the spherical interface, and r is the radius of curvature spherical interface (r > 0 if the interface appears convex to light impinging upon it from the -z direction, and r < 0 if the interface appears convex to light impinging upon it from the -z direction). You apply this formula twice: first to locate the image formed by the front surface of the sphere, and then to locate the image formed by the rear surface of the sphere (by considering the image formed by the front surface of the sphere to be the object that is imaged by the rear surface of the sphere).
Thanks a lot for the information. I was able to find the formula in the book (thanks to your description).
I'm having some doubts. I post a picture of the sketch of my draft. I must say that using $$\frac{n_1}{z_1}+\frac{n_0}{z_0}=\frac{n_1-n_0}{r}$$ for the convex part of the lens, my sketch is bad. I'm having doubt about $$z_1$$. I'm tempted to use it as 0.2 m, the diameter of the lens but it doesn't reach the right side of the equation which is worth 1/20.
And if I use $$z_1$$ as in my sketch, I reach a negative value for it, meaning that the image forms inside the lens, which I don't believe to be true.

I'd appreciate if someone could help me a bit more.

#### Attachments

las3rjock
To get you started, the equation for the first surface is
$$\frac{1.5}{z_1} - \frac{1}{-1.2} = \frac{1.5 - 1}{0.1}$$
which you should solve for z1. This gives a location for the image formed by the first surface relative to the vertex of the first surface. Then you treat that image as the object that is imaged by the second surface, and solve the appropriate equation for the second surface for z1.

Gold Member
To get you started, the equation for the first surface is
$$\frac{1.5}{z_1} - \frac{1}{-1.2} = \frac{1.5 - 1}{0.1}$$
which you should solve for z1. This gives a location for the image formed by the first surface relative to the vertex of the first surface. Then you treat that image as the object that is imaged by the second surface, and solve the appropriate equation for the second surface for z1.
Ok thanks. I realize I made a stupid error (I put 1/0.1=1/10 instead of 10).
However are you sure this is the right number for $$z_0$$? You wrote
where z0 is the displacement of the object from the spherical interface
You put $$z_0=1.2$$. So is it the distance between the center of the sphere and the object? I had used $$z_0=1.1$$. But I realize that for each light ray going from the object to the sphere, the distance is not the same. So why did you choose $$z_0=1.2$$ and not say 1.15 which might be the average distance the rays has to overcome in order to reach the sphere' surface? Just curious about the number 1.2m.

las3rjock
object situated at 1.2 m from a crystal ball
This part of the problem statement means that the object is located 1.2 m in front of the first refracting surface of the crystal ball (1.3 m from the center of the crystal ball, and 1.4 m from the second refracting surface of the crystal ball). The displacements z0 and z1 in the equation are measured from the refracting surface, not the center of curvature, therefore you use z0 = -1.2 m.

Gold Member
This part of the problem statement means that the object is located 1.2 m in front of the first refracting surface of the crystal ball (1.3 m from the center of the crystal ball, and 1.4 m from the second refracting surface of the crystal ball). The displacements z0 and z1 in the equation are measured from the refracting surface, not the center of curvature, therefore you use z0 = -1.2 m.

It makes sense, thank you very much!

Gold Member
I reach $$z_1=\frac{12}{105} \approx 0.1143 m$$. Is this situated at 0.1143 meter from the rear surface or from the front surface of the sphere? From the rear surface I'd bet.

Now I assume the object is situated at 0.1143 meter on the right side from the rear surface (so $$z_0=0.1143$$, $$n_0=1$$. $$z_1$$ is the unknown and $$n_1=1.5$$.) I believe I'm not getting it. I reach $$z_1=-\frac{18}{45}=-40 cm$$ from the rear surface. In other words, 40 cm at the right of the rear surface if you take my sketh as reference. Does this make sense that the image is closer to the sphere than the object is?