Mirror placed horizontally between two half convex lenses

[Null_Void]

Homework Statement

See pic

Relevant Equations

None

So I was able to find out quite easily the image distance after the first refraction and the corresponding magnification.

Employing Cartesian convention
$1/v - 1/u = 1/f$

Substituting the values gives us:
$v_1 = +60cm$
$m_1 = -3$

Thus the height of the image below the principal axis must be $6mm$

I was also able to find the horizontal position of the final image.
Now the problem arises when calculating the final image height for the refraction by the second half-lens.

The distance between the plane mirror and the principal axis is $4cm$ while the image is formed $6cm$ below the principal axis (height). I can see that $2cm$ of the image is formed as a virtual object and here's where I'm stuck.

1) Can the real image formed by the first lens on the plane mirror give rise to a virtual image of the same size?
This is my first time encountering a scenario like this and I'm a bit skeptical

2) how do I establish a relationship between the virtual object and the corresponding final image formed? I'm confused as there's a lot of stuff going on.

3) Will the virtual image formed by the plane mirror of the first real image, act as an object for the second lens? I feel that it shouldn't as it isn't a virtual object but the book solution seems to assume it does.

The height of the final image is given as 8/3mm

 

[kuruman]

I think it will be instructive for you to do part (b) first in order to get your bearings.

First make a ray diagram as if the mirror is not there. Try drawing it to scale as best as possible. As you have already found, you will get a real, inverted image at 60 cm from the first half-lens. Now put the mirror in place. Does it look like the image is still formed at 60 cm or does it look like the converging rays that would have formed the image at 60 cm diverge upon reflection from the mirror?

 

[haruspex]

It is very unclear what is meant by the "size" of the final image. The shown object consists of a 2mm long stick. As you found, only the top $\frac 43$mm form an image below the mirror, which folds that to be above the mirror. So the final image is of a folded stick. What is meant by its size?

 

[Null_Void]

I think it will be instructive for you to do part (b) first in order to get your bearings.

First make a ray diagram as if the mirror is not there. Try drawing it to scale as best as possible. As you have already found, you will get a real, inverted image at 60 cm from the first half-lens. Now put the mirror in place. Does it look like the image is still formed at 60 cm or does it look like the converging rays that would have formed the image at 60 cm diverge upon reflection from the mirror?

4mm of the object will be formed as a real image at 60cm, while the 2mm part will act as a virtual object, something like this:

So yes, the rays will definitely diverge after reflection from the plane mirror. I initially proceeded by considering the real image $A_1C_1$ and virtual object $C_1B_1$ to act as objects for the second lens but got the wrong answer. The book says that we have to consider the virtual image of $A_1C_1$ but why is that? How can a virtual image act as an object? It isn't a virtual object since rays do not converge toward it but rather diverge.

 

[kuruman]

The second lens does not care whether the rays that reach it are reflected off the mirror or emanate undeflected from a real object. As shown in the enhanced diagram below, reflected rays originating at the tip B of the object diverge from a common point of intersection. These rays are then brought into convergence and form a real image of point B to the right of lens 2. What is the horizontal distance from lens 2 that matters for finding the final image?

 

[Null_Void]

It is very unclear what is meant by the "size" of the final image. The shown object consists of a 2mm long stick. As you found, only the top $\frac 43$mm form an image below the mirror, which folds that to be above the mirror. So the final image is of a folded stick. What is meant by its size?

I think they mean the height of the image formed by lens-2. We are expected to find the magnification from the horizontal distance and then apply it to find the height of the final image

 

[Null_Void]

The second lens does not care whether the rays that reach it are reflected off the mirror or emanate undeflected from a real object. As shown in the enhanced diagram below, reflected rays originating at the tip B of the object diverge from a common point of intersection. These rays are then brought into convergence and form a real image of point B to the right of lens 2. What is the horizontal distance from lens 2 that matters for finding the final image?
View attachment 353784

It should be $60cm$ plus the distance to the imaginary intersection point of the reflected rays, but I'm not sure about how to find the required distance. Any hints?

 

[kuruman]

First imagine the mirror not being there. Also imagine a screen placed at 60 cm from the lens. A real inverted image of length 6 mm, i.e. three times the length of the object, will be seen on the screen.

Now imagine the mirror being slowly raised from below towards the optical axis. As long as the mirror is more than 6 mm below the optical axis, it might as well not be there. All the rays emitted in all directions from a single point on the object and pass through the lens will converge to a single point on the screen: there is a one-to-one correspondence from a point on the object, through the lens, to a point on the screen. When the mirror is exactly 6 mm from the axis, nothing will change and you will still a 6 mm image of the 2 mm object. It follows that if the object were 3 mm instead of 2 mm, you would see the bottom 2 mm on the screen and the top 1 mm will be chopped off.

That's because the rays emanating from the top 1 mm of the 3 mm object are reflected and diverge before reaching the screen; the one-to-one correspondence from a point on the object, through the lens, to a point on the screen is destroyed by the mirror for these points. The take home message is that if the mirror is closer to the optical axis than 3 times the length of the object, the image of the top part of the object is chopped off from appearing on the screen.

Finally, imagine removing the screen. Only the rays that would have converged on it to form the image, now diverge as if they emanated from a single point to reach lens 2. Therefore only part of the real image formed by lens 1 constitutes an object at distance (120 - 60 = 60) cm from lens 2.

 

[Null_Void]

Only the rays that would have converged on it to form the image, now diverge as if they emanated from a single point to reach lens 2. Therefore only part of the real image formed by lens 1 constitutes an object at distance (120 - 60 = 60) cm from lens 2.

But going by your logic, the object height will $4mm$ and the corresponding magnification at lens-2 will be $-1/3$. Therefore the final image height would be $4/3mm$, but the answer is given as $8/3mm$. Here's the solution:

 

[kuruman]

Yes, I incorrectly assumed, without proving it to myself, that intersection point B₂ in your solution does not exist. I now proved that it does exist. Sorry about the confusion.

It seems that you have a complete solution. Is there something about it that you don't understand?

 

[Null_Void]

Yes, I incorrectly assumed, without proving it to myself, that intersection point B₂ in your solution does not exist. I now proved that it does exist. Sorry about the confusion.

It seems that you have a complete solution. Is there something about it that you don't understand?

I'm very confused about the role of the plane mirror, so both, the part cut off by the plane mirror and the virtual image formed of the remaining part of the original real image act as a virtual object for the second lens?

 

[kuruman]

I'm very confused about the role of the plane mirror, so both, the part cut off by the plane mirror and the virtual image formed of the remaining part of the original real image act as a virtual object for the second lens?

To become unconfused, I think that you should set aside equations and virtual objects and concentrate on what it means to "see" an object such as the upright arrow AB on the far left of the figure in the solution. I hinted at that in post #8.

Suppose you are just looking at the arrow with no lenses or mirrors anywhere. Here is a list of propositions that are obvious in retrospect but must be stated before they come to the surface of one's thinking.

1. All points on the surface of the arrow are secondary sources of light, that is they reflect light from a source, e.g. a lightbulb. Proof: If you turn off the lightbulb you can no longer see the arrow.
2. Any one such point emits diverging rays of light in all directions. Proof: You can see the same point from many different angles as long as nothing obstructs the rays from it reaching your eye.
3. The diverging rays from the point reach the converging lens of your eye and focuses it at a point on your retina. The energy of the incoming photons is transduced by the rod and cone cells on your retina to an electrical pulse that your brain interprets as "seeing" the object. Proof: There is no simple proof. Look it up or ask an optometrist.

It follows that different pats of the object are focused on different parts on your retina but there a one-to-one correspondence between points on the screen and points on the retina, one and only one point on the object stimulates one and only point on the retina. All the rays from that point on the object converge on the same point on the retina.

Your brain is trained to interpret rays as traveling in straight lines with the object being at the intersection of the diverging rays that reach the lens of the eye. When you look into a mirror you see yourself behind the mirror even though you know perfectly well there is nobody there.

Now you are ready to understand what's going on with this problem. Imagine that the second half-lens on the right is the lens of your eye. Figure (c) in the solution shows what you would see, an arrow with its tail at A₂ and its tip at B₂. As far as lens 2 is concerned it is a real object because at line A₂B₂ light is emitted in all directions left to right.