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The reason is simple geometry. Consider the following image. http://images.tutorvista.com/content/light-reflection/plane-mirrors-inclined-90-image-formation.gif

Then you can clearly see the image at $O_{3}$ is due to double reflection from mirrors $M'$ and $M$. In general all the images including the object will be on a circle centered around the intersection point between two mirrors. For two mirrors inclined at $n^{0}$ with respect to each other.The number of images are $\frac{360}{n}-1$. I do not know the proof of the last formula. As far as I know it is because reflection of one mirror to another forms a virtual mirror and the number of virtual mirrors are $\frac{360}{n}$ ![] (http://www.learnquebec.ca/export/sites/learn/en/content/curriculum/mst/images/CHAPITRE523.GIF) Final image from both the mirror coincides hence we have to subtract $1$.

The reason is simple geometry. Consider the following image. [![][1]][1]
_{(source: tutorvista.com)}

Then you can clearly see the image at $O_{3}$ is due to double reflection from mirrors $M'$ and $M$. In general all the images including the object will be on a circle centered around the intersection point between two mirrors. For two mirrors inclined at $n^{0}$ with respect to each other.The number of images are $\frac{360}{n}-1$. I do not know the proof of the last formula. As far as I know it is because reflection of one mirror to another forms a virtual mirror and the number of virtual mirrors are $\frac{360}{n}$ ![] (http://www.learnquebec.ca/export/sites/learn/en/content/curriculum/mst/images/CHAPITRE523.GIF) Final image from both the mirror coincides hence we have to subtract $1$. [1]: https://i.sstatic.net/SVEGt.gif

The reason is simple geometry. Consider the following image. http://images.tutorvista.com/content/light-reflection/plane-mirrors-inclined-90-image-formation.gif

Then you can clearly see the image at $O_{3}$ is due to double reflection from mirrors $M'$ and $M$. In general all the images including the object will be on a circle centered around the intersection point between two mirrors. For two mirrors inclined at $n^{0}$ with respect to each other.The number of images are $\frac{360}{n}-1$. I do not know the proof of the last formula. As far as I know it is because reflection one mirror to another forms a virtual mirror and the number of virtual mirrors are $\frac{360}{n}$ ![] (http://www.learnquebec.ca/export/sites/learn/en/content/curriculum/mst/images/CHAPITRE523.GIF) Final image from both the mirror coincides hence we have to subtract $1$.

The reason is simple geometry. Consider the following image. http://images.tutorvista.com/content/light-reflection/plane-mirrors-inclined-90-image-formation.gif

Then you can clearly see the image at $O_{3}$ is due to double reflection from mirrors $M'$ and $M$. In general all the images including the object will be on a circle centered around the intersection point between two mirrors. For two mirrors inclined at $n^{0}$ with respect to each other.The number of images are $\frac{360}{n}-1$. I do not know the proof of the last formula. As far as I know it is because reflection of one mirror to another forms a virtual mirror and the number of virtual mirrors are $\frac{360}{n}$ ![] (http://www.learnquebec.ca/export/sites/learn/en/content/curriculum/mst/images/CHAPITRE523.GIF) Final image from both the mirror coincides hence we have to subtract $1$.

The reason is simple geometry. Consider the following image. http://images.tutorvista.com/content/light-reflection/plane-mirrors-inclined-90-image-formation.gif

Then you can clearly see the image at $O_{3}$ is due to double reflection from mirrors $M'$ and $M$. In general all the images including the object will be on a circle centered around the intersection point between two mirrors. For two mirrors inclined at $n^{0}$ with respect to each other.The number of images are $\frac{360}{n}-1$. I do not know the proof of the last formula.As far as I know it is because reflection one mirror to another forms a virtual mirror and the number of virtual mirrors are $\frac{360}{n}$ as final image from both the mirror coincides hence we have to subtract $1$. ![] (http://www.learnquebec.ca/export/sites/learn/en/content/curriculum/mst/images/CHAPITRE523.GIF)

The reason is simple geometry. Consider the following image. http://images.tutorvista.com/content/light-reflection/plane-mirrors-inclined-90-image-formation.gif

Then you can clearly see the image at $O_{3}$ is due to double reflection from mirrors $M'$ and $M$. In general all the images including the object will be on a circle centered around the intersection point between two mirrors. For two mirrors inclined at $n^{0}$ with respect to each other.The number of images are $\frac{360}{n}-1$. I do not know the proof of the last formula. As far as I know it is because reflection one mirror to another forms a virtual mirror and the number of virtual mirrors are $\frac{360}{n}$ ![] (http://www.learnquebec.ca/export/sites/learn/en/content/curriculum/mst/images/CHAPITRE523.GIF) Final image from both the mirror coincides hence we have to subtract $1$.