Problem of two honest buddies

This video is helpful for visualizing this problem and the equation $n = \frac{360°}{\theta°} - 1$: https://www.youtube.com/watch?v=6maFbmPCnzY&feature=youtu.be

If the angle were $45°$, the number of images would be $\frac{360°}{45°} - 1 = 7$, and if the angle were $51 \frac{1}{7}°$, the number of images would be $\frac{360°}{51 \frac{1}{7}°} - 1 = 6$. In this problem, since the angle $50°$ is between $45°$ and $51 \frac{1}{7}°$, there should be $6$ full images and $1$ partial image of the object.

The difference between the video David posted and the situation here: the posted video has an object that extends from mirror to mirror, but this problem has a point object.

So when the video shows that 7.2 objects are visible, these are: (a) the original object; (b) six complete images; (c) two partial images of only 1/10th of the object.

Our point object corresponds to the center of the larger object shown in the video. If we look on straight, we can see only six images. But turning our head left or right, we can catch sight of two more.

In my opinion, 6 images.

My way to deal with these cases is always the same: reflect the world around the mirrors, and reflect again, and again. In the drawing below, left side, you have the direct visible world inside the mirrors, and that one reflected a first time. The green arrow shows where the right mirror puts its first image. The right side shows the full schematic... after repeated reflections over the real mirrors (green), or a single reflection (in this case) over the image of each mirror produced by the previous first reflection operation (yellow).

This way, you don't need to concern in doing ray reflections... from source to eye-sight, the light can be made to trace a straight line. By mirroring image over the mirrors, your straight line passing through a mirror is equivalent to one that reflects using reflection law. There is however a cardinal rule: The reflections over the right side mirror can only be picked through the right side mirror, and the same for the left mirror. If as an observer, you are in the middle, seeing the images of an object in the middle, by symmetrical mirrors $\alpha$ apart, you'll see only images at $\pm \alpha$, $\pm 2\alpha$, $\pm 3\alpha$, until $\pm 180º$. For instance, in that situation, you cannot see an image at ${\color{#D61F06}X}$, because the light ray doesn't reach the eye through the right mirror (trace a ray and it doesn't cross it... things might be different with a different position for the observer, or if the mirror were semi-transparent and extended behind too).

From this, you can see the number of images seen is:

$2 \left \lfloor{\frac{180}{\alpha}}\right \rfloor$

...when $\alpha$ does not divides 180º, and that number minus 1, when it divides. In our case:

$2 \left \lfloor{\frac{180}{50}}\right \rfloor = 2\times 3=6$

PS.: I think I was too naive in my answer. When you asked how many images, I thought in images I can see... I'm not recalling well the definition of images in more formal contexts, nor will I try to recall it better, because I/you may have learn differently. But I thought I might be more precise in what I was meaning. First, I don't have any problem with what you did. You have a typo in an angle, it should be 100º where you have 110º, but beside that, I think I understand what you did and I have no qualms with that. Second, my 6 image guess is for an observer in the middle of the mirrors (A position). At that position, he'll see the original object and 6 images in the mirrors. But if he changes place, for the B position, he'll see 7 images.

Finally, C presents a scenario where we might be aware of 8 images at the same time.

An optical system changes the direction of light, in particular, the light emanated from sources of light we call objects. An image by an optical system is, for me, the point where we would have to put objects to get the same effect if we had not the optical system. This personal definition is probably my source of trouble with your problem, since it leads me to a different number of images depending on the position of an observer. If however we are talking about the full of positions where images can be, then I must agree with the number 8 of your and other answers. The way to prove it is by making a lens big enough to capture all the light between the mirrors, and project it in images after. Then, we'll discover there are 8 images.

By the way, I did the actual experiment with two mirrored doors in my bathroom, making 50º between them. The results agreed with what I said.

@Mahdi Raza, @David Vreken, @Vinayak Srivastava, @Ram Mohith, @Hamza Anushath, @jordi curto, @Vikram Karki, @Stef Smith, @Alak Bhattacharya, @Páll Márton, @Maria Paszkiewicz, @Finnley Paolella, @Hana Wehbi, @Yajat Shamji, @Richard Desper, @Pi Han Goh

Hello Vikram ,

The formula a think is Integer ( 360/50) -1 . That mean on produce 6 images.

I did'n prove that yet , but I will explain way I would do:

Graphicaly all posible ray starting at point has to retorn same point , First ray is that perpendicular at mirror right .Second ray 25º to normal mirror right and reflect to left mirror and for simetry retorn to point . Third ray is 50 º to normal right mirror reflect perpendicular to left mirror and retorn same path . This three ray has 3 more symetric on left mirror , total 6 rays returning to point , that implie 6 images , the ray going from point to vertice not retorn to point then not produice imagen .

PD: going on explanation : We can observe that over ski line (half right) ,angles that alow ray to retorn to point object are 25 , 50 and 75 next will be 100 but is greather than 90 and ray loose . Then number of good angles are Integer ( 90 / (50/2) ) and generalising with any angle between mirrors named "A" , number good angles right side = Integer(180/A) . By simetry , the same for left side ,we multiply by 2 , but if ray with 90º exist , we repeat that (case 180/A = integer) and we must sustract one repetition . Then formula will be : number of images (not including object) = 2 x (integer(180/A) - 1 x (180/A = integer)

Here is drawn only three images , by simmetry on can draw other three . Scale not correct :

• With $50^{\circ}$ as the angle between the two plane mirrors, it is logical to divide that by the total number of degrees $= 360^{\circ}$.
• $\dfrac{360}{50} = 7.2$, a fractional number means a partial image. Thus we will get 7 copies
• But remember that one of them is the original object. Thus we might \boxed{6} and some partial images

References: Quora answer, Video as mentioned by @David Vreken

I assume that the object is a point, infinitely small; so you cannot produce fractional images. The only situation in which you get an odd number of images is when the mirror image lies precisely on the line of symmetry. But that cannot be the case here.

It can be shown (I'm sure the video does this, too) that the image points and object lie on a circle, at an arc distance of 50˚ from each other. Thus, if we call the object's position 0˚, and the mirrors cross the circle in ±25˚, then the eight images will be located at ±50˚, ±100˚, ±150˚, ±200˚ (= ±160˚). These last two images will not be reflected further, because they lie on the wrong side of the planes of the mirrors.

In general, if the mirrors form angle $\alpha$, the mirror images will be located at $\pm \alpha, \pm 2\alpha, \pm 3\alpha, \dots$ up to, and including the first value $\pm n\alpha$ where $n\alpha \geq 180˚$. This suggests that the number of images is $2\times \left\lceil\frac{180^\circ}{\alpha}\right\rceil,$ which is equal to $360^\circ/\alpha$ rounded up to the next even integer. In case $\alpha | 180^\circ$, the last image will lie precisely on the line of symmetry and we get an odd number of images, namely $360^\circ/\alpha + 1$.

In this case, we get $360^\circ/50^\circ = 7.2$ which we round up to the next even integer, that is, 8.

It think the answer is that eight images are formed.

I am not sure, and I don't know whether I am right or not

I don't know these type of questions

-EDIT-

I GOT THE ANSWER

The formula I used is $\frac{360°}{x°}$ (if and only $\frac{360°}{x°}$ is odd, if it is even, we use $\frac{360°}{x°} - 1$ )

$\frac{360°}{50°} = 7.2$

As seven is odd, and it is approximately near to the answer we got, $7.2$, we need not subtract, and there are in total approximately seven images formed.

It might be safe to say that seven images are formed

I only understand the first pair of images, don't know how you progress after that.

I would argue that the number of images should be even, because of symmetry.

Your picture is incorrect. The second image should be $100^{\circ}$

In my Physics Online Class of this year, I was told the same formula given by @David Vreken and @Mahdi Raza, but it is also told that the no. of images is always odd, I don't know why.

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