Image if a mirror is placed at the focal distance of a biconvex lens

[HastiM]

Hello,

suppose we are given a biconvex lens with focal length of 4 cm. Further, an arrow of height 1 cm is placed in front of the lens in a distance of 8 cm (i.e. twice the focal length). Lastly, a mirror is placed on the other side of the lens at a distance of 4 cm and such that it performs an angle of 45 degrees with the optical axis and such that no reflected ray will hit the lens again. I want to know where the image of the arrow will occur.

Of course, if the mirror would not exist, then the image of the arrow would appear upside down at a distance of 8 cm. But I am very confused because of the mirror. My problem is that some of the rays emitted by the arrow become parallel after a reflection at the mirror. At the same time, some rays coming from the tip of the arrow intersect at several points after passing through the lens. Could someone please help me with the present situation?

I would very appreciate your help

 

[Ibix]

Have you drawn a ray diagram? If you post it we may be able to see where you went wrong.

What do you expect a 45° mirror to do?

 

[HastiM]

Hello Ibix,

thank you for your answer. I have drawn a ray diagram (see the picture which I have uploaded).

As I said, I really do not know what happens due to the mirror. Do you have any hint for me?

 

[HastiM]

Another thing which bothers me is that light rays, which are emitted by the arrow, may intersect twice after passing through the lens, as it is seen in the picture (if the picture is correct of course). I am wondering, where is the image point then? Both points or only one of them…?

 

[Ibix]

The diagram looks fine - except you need to use a ruler and measure distances and angles precisely. Then the picture will tell you precisely what happens.

As I said, I really do not know what happens due to the mirror. Do you have any hint for me?

Well, what do mirrors do? If you look at a 45° mirror where do you end up looking? Is there anything special about light rays that happen to enter your eyes?

Another thing which bothers me is that light rays, which are emitted by the arrow, may intersect twice after passing through the lens, as it is seen in the picture (if the picture is correct of course). I am wondering, where is the image point then? Both points or only one of them…?

A good question. The point about the image is that it occurs where all rays emitted from the object cross. You can relatively easily add another ray from the tip of the arrow in a straight line through the centre of the lens. If you've drawn it precisely (!) this should go through the upper crossing point, but not through the lower one.

 

[Drakkith]

My problem is that some of the rays emitted by the arrow become parallel after a reflection at the mirror.

That shouldn't happen. In your diagram you have 2 rays leaving the arrow, but 3 rays reflecting off of the mirror, two of which reflect from the same point on the mirror but reflect in different directions. Where did this extra ray come from?

 

[Ibix]

That shouldn't happen. In your diagram you have 2 rays leaving the arrow, but 3 rays reflecting off of the mirror, two of which reflect from the same point on the mirror but reflect in different directions. Where did this extra ray come from?

I read the right hand vertical line as a reflection of the optic axis, which is fine. If it was intended as a ray then, as Drakkith says, it shouldn't be there.

 

[Drakkith]

Another thing which bothers me is that light rays, which are emitted by the arrow, may intersect twice after passing through the lens, as it is seen in the picture (if the picture is correct of course). I am wondering, where is the image point then? Both points or only one of them…?

Take your finger and trace both rays at the same time at the same speed. The image location should approximately be where your fingers finally touch. I say "approximately" because your fingers aren't sharp and narrow instruments like a pen is. Also, if you draw more rays the focal point is the place where they all converge (ignoring aberrations), and you'll find that there's only one point for this.

 

[HastiM]

I want to thank both of you very much! I think that I now have a better understanding of what is happening :-). Best regards