Solve Width of Aperture: Find Width Now!

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To find the width of the aperture, apply the principles of similar triangles using the given measurements. A point source of light creates a bright patch 10.2 cm wide on a screen 1.02 m behind the aperture, which is 2.3 m away. By drawing a diagram, the relationship between the distances and widths can be established. The solution involves basic geometry rather than complex physics concepts. A clear understanding of similar triangles will lead to the correct answer.
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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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