Solve Width of Aperture: Find Width Now!

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SUMMARY

The discussion centers on calculating the width of an aperture based on the geometry of similar triangles. Given a point source of light illuminating an aperture 2.3 meters away, and a 10.2 cm-wide bright patch appearing on a screen 1.02 meters behind the aperture, the solution involves applying basic geometric principles. Participants emphasize the importance of drawing a diagram to visualize the problem and facilitate the calculation of the aperture's width.

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  • Understanding of similar triangles in geometry
  • Basic knowledge of light propagation
  • Ability to draw and interpret geometric diagrams
  • Familiarity with basic measurement units (meters and centimeters)
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Students studying geometry or physics, educators teaching optics, and anyone interested in understanding the relationship between light and geometric shapes.

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Width of Aperture...please help!

Homework Statement


A point source of light illuminates an aperture 2.3 m away. A 10.2-cm-wide bright patch of light appears on a screen 1.02 m behind the aperture. How wide is the aperture?

Homework Equations





The Attempt at a Solution


I can't figure out how to do this?
 
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woops this was supposed to go under physics...please remove! thanks!
 
… just geometry …

BuBbLeS01 said:

Homework Statement


A point source of light illuminates an aperture 2.3 m away. A 10.2-cm-wide bright patch of light appears on a screen 1.02 m behind the aperture. How wide is the aperture?

Actually, this is geometry, not physics.

Just draw a diagram, apply ordinary geometry of similar triangles, and you should get the answer! :smile:
 

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