Lens Maker's Formula - Confusion in the derivation

[Null_Void]

Homework Statement

Minor confusion in the derivation of the lens maker's formula

Relevant Equations

None

I was looking at the derivation of the lens maker's formula and I have a minor confusion which does not seem to go away.

So it is derived from the same principle for the refraction of a light ray at a spherical convex/concave surface, except that it undergoes refraction twice due to two surfaces (of the convex lens).

So here's what I understood:

The rays from the object are initially refracted at the first surface and forms a virtual image (am I right here?)to the right of the lens, and this image now acts as a virtual object for the final real image to be formed by the second surface of the lens after the second refraction.



For simplicity let's assume that the medium to the right and the left of the lens are same.
Let mediums be $n_1$ and $n_2$

Now I understand the equation for the first refraction.
The rays are from the left coming from medium $n_1$ and refract into medium $n_2$ and thus the equation is:

$n_1/u + n_2/v_1 = \frac {n_2 - n_1} {r_1}$

Now at the second refraction, rays from the virtual object strike the surface to the right of the optical center, refract and thus form the final image.

The second equation is given as:
$n_2/v_1 + n_1/v = \frac {n_1-n_2} {r_2}$

My doubt is, for the second refraction, the rays from the virtual object are from medium $n_1$ into medium $n_2$ (from the right side of the lens) like for the first refraction
Then shouldn't the equation be something like:

$n_1/v_1 + n_2/v = \frac {n_2 - n_1} { r_2}$

Why are the refractive indices changed for the subsequent equations? This has been bugging me for a while now and any help is greatly appreciated

Thanks in advance!

 

[haruspex]

To get the second situation from the first you need to:

• swap $n_1, n_2$
• substitute $-v_1, -v$ for $u, v_1$
• change $r$ to $-r$

Doesn’t that give the right result?

 

[Null_Void]

To get the second situation from the first you need to:

• swap $n_1, n_2$
• substitute $-v_1, -v$ for $u, v_1$
• change $r$ to $-r$

Doesn’t that give the right result?

But why do we swap the refractive indices, like I've mentioned in the post, both refractions seem to be the same case albeit from different directions. For both refractions, the rays are from $n_1$ to $n_2$

 

[haruspex]

But why do we swap the refractive indices, like I've mentioned in the post, both refractions seem to be the same case albeit from different directions. For both refractions, the rays are from $n_1$ to $n_2$

There are two ways to think of the second stage:

1. As rays from the first image to the final image. In this case it is from $n_2$ to $n_1$. The substitutions are as in post #2.
2. As rays from the final image to the first image. In this case, just substitute v for u and -r for r.

 

[Null_Void]

There are two ways to think of the second stage:

1. As rays from the first image to the final image. In this case it is from $n_2$ to $n_1$. The substitutions are as in post #2.
2. As rays from the final image to the first image. In this case, just substitute v for u and -r for r.

I'm sorry I still don't get it. In both cases the rays are still from $n_1$ into $n_2$. The virtual object is formed in the medium $n_1$, same for the final image. So both rays travel from $n_1$ to $n_2$ right? Or am I missing something?

 

[haruspex]

I'm sorry I still don't get it. In both cases the rays are still from $n_1$ into $n_2$. The virtual object is formed in the medium $n_1$,

No. The formula for the first stage only considers the ray entering the glass. Therefore the image is, theoretically, inside glass.

 

[Null_Void]

No. The formula for the first stage only considers the ray entering the glass. Therefore the image is, theoretically, inside glass.

So, it's like looking through a mirror immersed in a medium. The image formed will also appear to be placed in the same medium?

But now another confusion arises, so the first surface, second surface and the virtual object are all in the same medium, then how will the rays refract at the second surface? Shouldn't they not be affected?
How do I draw a ray diagram depicting rays travelling from the virtual object and refracting at the second surface?

Will the rays from the virtual object retrace the dotted lines and after refraction travel in the opposite direction to the direction in the which the actual image is formed?

 

[haruspex]

So, it's like looking through a mirror immersed in a medium. The image formed will also appear to be placed in the same medium?

When applying math to physics it is often important to be aware of what exactly your equations are describing. The equation for the first refraction says nothing about a second surface, so it tells you where the image would be if everything after entering the glass is just more glass.

But now another confusion arises, so the first surface, second surface and the virtual object are all in the same medium, then how will the rays refract at the second surface? Shouldn't they not be affected?

Not sure what you mean by a surface being "in" a medium. It is a boundary between media.

How do I draw a ray diagram depicting rays travelling from the virtual object and refracting at the second surface?

Will the rays from the virtual object retrace the dotted lines and after refraction travel in the opposite direction to the direction in the which the actual image is formed?

The second way of looking at it that I mentioned in post #4 may help. Rays follow the same path if you reverse them, so instead of following rays from the first image to the final image, you can ask where you would place the final image to produce an image where the first image is.
The final image, $I_2$, we know, is in air, somewhere to the right of the lens. (Reversed) rays from there go leftwards to enter the lens from the right, so this is an air to glass transition. Just as we ignored the right side of the lens in the first equation, we now ignore the left side. This means the observer is in the glass and perceives an image ($I_1$) as though it were also in glass. That is consistent with how the first equation treats $I_1$.

 

[Null_Void]

When applying math to physics it is often important to be aware of what exactly your equations are describing. The equation for the first refraction says nothing about a second surface, so it tells you where the image would be if everything after entering the glass is just more glass.

So If an observer were positioned at the left of the lens, then the first image is what he would see as a virtual image, right? Is it analogous to seeing a virtual image through, say in a plane mirror?

The second way of looking at it that I mentioned in post #4 may help. Rays follow the same path if you reverse them, so instead of following rays from the first image to the final image, you can ask where you would place the final image to produce an image where the first image is.

When you put it like that, it makes complete sense and the result is apparent from the ray diagrams. But I still can't visualize the first case. Do the rays travel from the virtual object towards the second surface? How are virtual objects capable of producing images if there are no rays emerging from them?

I can see that the actual rays must refract twice. But when we use the virtual object approach, if the virtual object is in medium $n_2$, then why should it refract at the second surface of the lens, since it is of the same medium?
I'm confused about image formation by virtual objects in general.

 

[Steve4Physics]

I'm confused about image formation by virtual objects in general.

Can I chip in? The terminology 'virtual object' may be the source of confusion.

(From https://physics.stackexchange.com/q...rence-between-real-and-virtual-objects-optics)

Real objects are points from which light rays diverge.
Virtual objects are points towards which light rays converge.
(You might find that counter-intuitive, but that's the accepted terminology.)

(Real images are points towards which light rays converge.
Virtual images are points from which light rays diverge.)

There's even a Khan Academy video exaplaining virtual objects:

 

[Null_Void]

Can I chip in? The terminology 'virtual object' may be the source of confusion.
View attachment 353740
(From https://physics.stackexchange.com/q...rence-between-real-and-virtual-objects-optics)

Real objects are points from which light rays diverge.
Virtual objects are points towards which light rays converge.
(You might find that counter-intuitive, but that's the accepted terminology.)

(Real images are points towards which light rays converge.
Virtual images are points from which light rays diverge.)

There's even a Khan Academy video exaplaining virtual objects:

@Steve4Physics thanks for the resources, they were very helpful. But if you don't mind, I have certain questions from the video.

In the end we have a double lens system and we are to find where the final image will be formed and naturally we use the concept of a virtual object to do so.
My question is, does a virtual object carry the same properties as a real object? since in the video it is mentioned that we assume light rays to emerge from the virtual object.

Will the rays from the virtual object refract in the exact opposite direction to the direction of the rays in which the original final image is formed?

 

[Steve4Physics]

In the end we have a double lens system and we are to find where the final image will be formed and naturally we use the concept of a virtual object to do so.
My question is, does a virtual object carry the same properties as a real object? since in the video it is mentioned that we assume light rays to emerge from the virtual object.

Will the rays from the virtual object refract in the exact opposite direction to the direction of the rays in which the original final image is formed?

I didn't watch the whole video so can't answer your questions. When I get a chance I'll watch it and might be able to answer.

 

[Steve4Physics]

With reference to the Post #11 video:

In the end we have a double lens system and we are to find where the final image will be formed and naturally we use the concept of a virtual object to do so.
My question is, does a virtual object carry the same properties as a real object?

It depends what you mean by 'properties'. The simple answer is 'no' - or they would be the same!

Light rays emerge (diverging) from a real object. This is not true of virtual objects.

A virtual object is just a convenient concept when analysing multi-element systems. It corresponds to the position where element number N would have formed an image if element number N+1 hadn't prevented it. This is then the position of the object of element N+1.

since in the video it is mentioned that we assume light rays to emerge from the virtual object.

I didn't see that. Can you give the time into the video? As far as I'm aware it's wrong.

Will the rays from the virtual object refract in the exact opposite direction to the direction of the rays in which the original final image is formed?

There are no rays from a virtual object!

 

[Null_Void]

@Steve4Physics First of all, thank you for taking the time to go through the video.

didn't see that. Can you give the time into the video? As far as I'm aware it's wrong.

I'm sorry that was extremely misleading, the video said that the rays appear to converge.

I understand the need for a virtual object, but I still don't understand the image formation by a virtual object. Do we have specifics for how an image is formed by a virtual object?

There are no rays from a virtual object

But then how would you consider image formation?

 

[Steve4Physics]

There are no rays from a virtual object!

But then how would you consider image formation?

Try this.
You have an optical component, C, with incident rays travelling left to right. Suppose C's incident rays converge at point P situated on the right of C.

We say P is the virtual object for C . Think of P as attracting the incident rays.

A virtual object acts as if it is ‘pulling in’ C’s incident rays. (Compare that to a real object.)

 

[Null_Void]

Try this.
You have an optical component, C, with incident rays travelling left to right. Suppose C's incident rays converge at point P situated on the right of C.
View attachment 353787
We say P is the virtual object for C . Think of P as attracting the incident rays.

A virtual object acts as if it is ‘pulling in’ C’s incident rays. (Compare that to a real object.)

I think I get it now. I believe my confusion was due to the assumption that virtual objects act like real objects. Nevertheless, your posts cleared it up.

Thank you @haruspex and @Steve4Physics for your time and patience!