What is the angle of the the mirror?

  • Thread starter Thread starter InSpiRatioNy
  • Start date Start date
  • Tags Tags
    Angle Mirror
InSpiRatioNy
Messages
37
Reaction score
0
[Urgent] What is the angle of the the mirror?

Homework Statement


music-lesson.jpg

There is an intriguing aspect to Vermerr's "The Music" painting. If you notice, in the mirror there is a reflection of the woman and above that there is a reflection of what appears to be the feet of an easel. Upon closer inspection you can see the base of the back wall reflected in the mirror. The intersection of the back wal and floor in the mirror is marked on the outline in the image. We can use this as a second method to determine the length of the room. But the question is, what is the angle from the wall at which the mirror hangs?
Picture given: http://www.freeimagehosting.net/image.php?2e866c251b.jpg

Side view of the room (for measuring the angle):

http://img505.imageshack.us/img505/2651/musiclesson2xx1.jpg
http://g.imageshack.us/img505/musiclesson2xx1.jpg/1/

Homework Equations


Don't Know


The Attempt at a Solution


I've calculated the length of the room: 6.6 m but it might be wrong, still just base it on 6.6. meters.

This is already the last part, so help if you can! Thanks!
 
Last edited by a moderator:
Physics news on Phys.org


Do you see the dark shadow and lighter shadow at the paintings and the open chests right side ?? The long and lighter shadow is caused by the first set of windows. The windows at back can reach where the front ones cannot so that makes it lighter. As you can see there is a darker shadow also. I suppose that the room consists of 2 sets of 4 windows according to dark and light shadows and my predictions. Because if there were more sets of windows then there would be dark, light and lighter shadows. So by calculating the length of the dark shadows (only dark ones) at the time where they are equal by lenght, it shows that, at that point they are equally far from the front wall. Aquire the lenghts of painting and chest, desk or whatever it is, you can calculate the angle. but maybe the colomn in the middle is reflection of room. I mean if there is 3 metres front the middle colomn there will be 3 at back so length is 3+3+lenght of colomn.
 


So my length is probably about right.

What confuses me is how to get the angle... I think it may have something to do with optics on mirrors... I know the angle should be 15 degrees or about that because I measured it from a to scale diagram, but I need to actually find it...
 


since it's a painting the "actuality" is not even concerned. But try the dark shadow thing i said right up.



edit: whatever you do with the painting, you will get a ratio of lenghts, so how can you predict the actuall? The lady may be 1,50m or 1,80m and the rooms length would slightly change. I suppose you need something like a tiles lengh.
 
Last edited:


okay

Thanks
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top