Number of images formed by 3 mutually perpendicular plane mirrors

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SUMMARY

The discussion centers on the number of images formed by three mutually perpendicular plane mirrors in a rectangular room. An observer sees a total of 6 images due to reflections from the three planes, including additional reflections from the edges and the corner of the mirrors. The calculations for image formation by a convex mirror with a radius of curvature of 60 cm are also addressed, confirming that the image is virtual and located 15 cm in front of the mirror. Additionally, the relationship between object distance and magnification for a convex mirror is established, leading to the formula u = (1-n)f.

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  • Understanding of geometric optics, specifically image formation by mirrors.
  • Familiarity with the mirror formula: 1/u + 1/v = 1/f.
  • Knowledge of magnification concepts in optics.
  • Basic skills in ray tracing techniques for visualizing light paths.
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  • Learn about the properties of convex mirrors and their applications in optical systems.
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  • Investigate the implications of sign conventions in optics for accurate calculations.
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Amith2006
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Sir/Madam,
1)The 2 adjacent walls and the ceiling of a rectangular room are mirror surfaced.What is the number of images of himself does an observer sees?
If it was the case of 2 mirrors I know it is 3. But here there are 3 mirrors which are mutually perpendicular. But if we take 2 mirrors at a time then the total number of images is 9. What do you say Sir/Madam?
2)A convergent beam of light is incident on a convex mirror of radius of curvature 60 cm as shown in figure. What is the nature and position of the image formed by it?
I solved it in the following way:
Let u, v and f be the object distance, image distance and focal length of the convex mirror respectively.
Here u = + 10 cm, f = +30 cm
1/u + 1/v = 1/f
v = (uf)/(u-f)
= (10 x 30)/(10 – 30)
= -15 cm
Hence the image formed is virtual and 15 cm in front of the mirror. Is it right?
3)A convex mirror of focal length f produces an image (1/n)th of the size of the object. What is the distance of the object from the mirror?
I solved it in the following way:
Magnification = 1/n = -(v/u)
i.e. v = -(u/n)
1/u + 1/v = 1/f
1/u – (n/u) = 1/f
By solving I get,
u = (1-n)f
Are my sign conventions right? Sometimes the diagram may not be clear. So I will try to describe the diagram. A convergent beam of light serves as a virtual object which appears to converge at a distance of 10 cm behind the mirror.
 

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On question -1-, I would guess that you get one additional image at the corner of the 3 planes. Since it is a cube corner, you should get a reflection from it back to you in the middle of the room. I don't know if it's a real image, though. Have to try it or else do some ray tracing sketching to figure that out. So you'd get the 3 reflections from the planes, the 2 reflections from the 2 edges, and the 1 from the corner. Is a total of 6 the correct answer for question -1-?
 

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