Determining light intensity with a mirror

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Homework Help Overview

The discussion revolves around determining how the intensity of light at a screen is affected by the introduction of a mirror placed behind a point light source. The problem involves concepts related to light intensity, distance, and the behavior of virtual images.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between distance and intensity, questioning how the mirror creates a secondary source of light. There are discussions about the effects of distance on intensity and attempts to clarify the distances involved in the setup.

Discussion Status

The conversation is ongoing, with participants actively questioning assumptions and clarifying concepts related to the distances of the sources and their impact on intensity. Some guidance has been provided regarding the relationship between distance and intensity, but no consensus has been reached on the final outcome.

Contextual Notes

Participants are grappling with the concept of virtual images and their implications for calculating intensity. There is a focus on understanding how the distances from the source to the screen and the mirror to the screen affect the overall intensity observed.

digitaljeff
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Homework Statement



You put a point source of light (S) a distance (d) in front of screen (A). How is the light intensity at the center of the screen changed if you put a completely reflecting mirror (M) a distance (d) behind the source?

M-----d-----S-----d-----A

Homework Equations



I=Power/Area ??

The Attempt at a Solution



The answer it is giving in the book is 10/9 of the original intensity but i have no clue at all how to get this.
 
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When you put the mirror in place, the image constitutes a second source in addition to the original one. How far from the screen is this second source? How does intensity depend on distance?
 
i would think intensity in this case would be power/distance, so S/d originally. then with the mirror S/d + S/2d ... am i on the right track??
 
No. Answer me this, if you double the distance between the source and the screen, by what factor is the intensity on the screen reduced?
 
kuruman said:
No. Answer me this, if you double the distance between the source and the screen, by what factor is the intensity on the screen reduced?

decreases by the square of the distance from the source?? i think.. just re-reading my textbook trying to figure it out
 
digitaljeff said:
decreases by the square of the distance from the source?? i think.. just re-reading my textbook trying to figure it out
Correct. Now call the original screen-to-source distance one "unit". How many "units" is the image-to-source distance?
 
image to source would be 2 units, so intensity would be decreased by a factor of 4 for the light coming from the image on the screen compared to light coming from the source on the screen? .. argh sorry if I am clueless..
 
Two units is the mirror-to-source distance. The image is behind the mirror by an additional how many units?
 
kuruman said:
Two units is the mirror-to-source distance. The image is behind the mirror by an additional how many units?

argh forgot about the whole "virtual" image thing..

virtual image to screen would be 3 units ... as source to mirror is 1 unit.. then virtual image to mirror another 1 unit.. then 1 unit to screen..

sooo..

3 units from virtual image to screen... 1 unit from source to screen..?
 
  • #10
Correct. So if the intensity of the primary source to the screen is "1" intensity unit, what is the intensity of the secondary source that is 3 distance units away?
 

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