- #1
Amith2006
- 427
- 2
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.
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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