Determining light intensity with a mirror

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SUMMARY

The discussion focuses on determining the light intensity at the center of a screen when a completely reflecting mirror is placed behind a point light source. The original intensity is affected by the introduction of a virtual image created by the mirror, resulting in a total intensity of 10/9 of the original intensity. Key calculations involve understanding that the distance from the virtual image to the screen is three units, while the distance from the primary source to the screen is one unit. The intensity from the virtual image decreases by a factor of four due to the inverse square law of light intensity.

PREREQUISITES
  • Understanding of light intensity and the inverse square law
  • Familiarity with the concept of virtual images in optics
  • Basic knowledge of point sources of light
  • Ability to perform calculations involving distance and intensity
NEXT STEPS
  • Study the inverse square law of light intensity in detail
  • Learn about virtual images and their properties in optics
  • Explore the effects of mirrors on light propagation
  • Investigate practical applications of light intensity calculations in physics
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding the behavior of light and its intensity in various configurations.

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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