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Homework Statement
How much of wall 3m behind you can be observed in a 5cmsquare mirror which is held centrally at a distance of 10cm to your eye?
Homework Equations
N/A
The Attempt at a Solution
I see no way of doing this
Mirror Question: How Much of Wall 3m Behind You Can be Seen?
- Thread starter Nicholas.D
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How much of wall 3m behind you can be observed in a 5cmsquare mirror which is held centrally at a distance of 10cm to your eye?
N/A
I see no way of doing this
- #2
tiny-tim
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Welcome to PF!
Hi Nicholas! Welcome to PF!
Hint: where is the image? think of the mirror as a frame though which you can see part of the image of the wall (on the other side of the mirror)!
(btw, the same principle applies to a hologram … if you break a hologram in pieces, you can still pick up one piece and see the whole image, but only if you move your eye from side to side, to keep changing that bit of the image which is "in the frame")
Hi Nicholas! Welcome to PF!
Hint: where is the image? think of the mirror as a frame though which you can see part of the image of the wall (on the other side of the mirror)!
(btw, the same principle applies to a hologram … if you break a hologram in pieces, you can still pick up one piece and see the whole image, but only if you move your eye from side to side, to keep changing that bit of the image which is "in the frame")
- #3
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hmm ok I did try thinking about it like that. I also then tried to draw a few ray diagrams to help, but couldn't really get them to work. The answer uses similar triangles, but again, I couldn't make sense of why it works. If you have a strait mirror that doesn't focus the light onto a focal point, then why do you get magnification apparent in this particuler case?
Thanks a heap
Thanks a heap
- #4
tiny-tim
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Hi Nicholas!
You don't get magnification.
Nicholas.D said:
You don't get magnification.
Have you drawn it yet, with the image beyond the mirror?
What does your diagram look like?
What does your diagram look like?
- #5
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Yeah I did try and draw the set up.
To the right of my page I had a vertical line to represent the wall, then in the centre I had a smaller one to represent the mirror. I then tried to draw in a few rays from the top of the wall to the mirror. The only rays however that would form an image able to be seen by the viewer come in from the section of the wall directly in front of the position on the mirror, in other words, using my diagram, a 5cm^2 section of the wall would be able to be seen, but this is not the case.
Where is my logic flawed?
To the right of my page I had a vertical line to represent the wall, then in the centre I had a smaller one to represent the mirror. I then tried to draw in a few rays from the top of the wall to the mirror. The only rays however that would form an image able to be seen by the viewer come in from the section of the wall directly in front of the position on the mirror, in other words, using my diagram, a 5cm^2 section of the wall would be able to be seen, but this is not the case.
Where is my logic flawed?
- #6
tiny-tim
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Nicholas.D said:
Sorry, this makes no sense
…the wall (the imaginary one that you've drawn beyond the mirror) is the image …
the only question is how much of it can be seen from your eye.
btw your eye should also be in the drawing …
perhaps if you put it in, and draw rays from it, the answer will be obvious?
perhaps if you put it in, and draw rays from it, the answer will be obvious?
- #7
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won't the image seen by my eye just be the 5cm^2 portion of the wall reflected by the mirror? I'm having trouble understanding how a section of the wall larger than 5cm^2 can be seen in the mirror if there isn't any magnification
- #8
tiny-tim
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forget about reflection!
your diagram should not show any reflection (ie bent lines), it should only show straight lines …
from the eye, through the mirror, to the image of the wall
your diagram should not show any reflection (ie bent lines), it should only show straight lines …
from the eye, through the mirror, to the image of the wall
- #9
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I see now, and I think I understand just what to do.
My maths looks like this:
2.2/10=x/310 ie I used similar triangles to work out how much of the image of the wall I could see. However, the answers use 300 as the distance from the mirror to the image of the wall and 2.5/10 instead of 2.2/10. (I just rooted the 5 to get the side length of the mirror). Have I made another mistake?
Thanks for all your help btw :)
My maths looks like this:
2.2/10=x/310 ie I used similar triangles to work out how much of the image of the wall I could see. However, the answers use 300 as the distance from the mirror to the image of the wall and 2.5/10 instead of 2.2/10. (I just rooted the 5 to get the side length of the mirror). Have I made another mistake?
Thanks for all your help btw :)
- #10
tiny-tim
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Hi Nicholas!
(just got up :zzz: …)
Yes, you obviously have the right idea now!
Your 310/10 is correct … if the book uses 300, it's being lazy.
But I don't understand either your √5 or the book's 2.5 … perhaps I'm misunderstanding the question … what exactly does the question ask for?
(and is the mirror 5 cm2, or a 5 cm square?)
(just got up :zzz: …)
Nicholas.D said:
Yes, you obviously have the right idea now!
Your 310/10 is correct … if the book uses 300, it's being lazy.
But I don't understand either your √5 or the book's 2.5 … perhaps I'm misunderstanding the question … what exactly does the question ask for?
(and is the mirror 5 cm2, or a 5 cm square?)