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 $I_1$.