Mastering the Mirror Equation for Concave Mirrors in Physics 11

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Homework Help Overview

The discussion revolves around applying the mirror equation to determine the image location and height for an object placed at the center of curvature of a concave mirror. Participants are exploring the relationships between focal length, object distance, and image characteristics in the context of concave mirrors.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to clarify the mirror equation and its application, questioning the relationship between the center of curvature and the focal point. There are discussions about the definitions of focal length and object distance, with some participants expressing confusion about how these concepts relate to the problem.

Discussion Status

The discussion is ongoing, with participants providing hints and corrections regarding the relationships involved. Some guidance has been offered on substituting values into the mirror equation, but there is no explicit consensus on the solution or approach yet.

Contextual Notes

There appears to be some confusion regarding the definitions and relationships of focal length and object distance, which may affect the problem-solving process. Participants are also navigating the constraints of homework expectations and the need for clarity in their understanding.

leehom
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Use the mirror equation to find the image location and the height of an object placed at the centre of curvature of a concave mirror. Also find the magnification. Hint: What is the relation between the focal length and the object distance, do, for this situation?

I'm really confused about this question... Can someone please help me?
 
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What's the mirror equation? (It's the same as the thin-lens equation)

What's the relationship between the center of curvature and the focal point?
 
centre of curvature is half the focal point and the mirror equation is 1/di + 1/do = 1/f
 
but i still don't understand how that answers the question..
 
I was answering your other thread.

Center of curvature is not half the focal length, it's the other way 'round.

Center of curvature is twice the focal length. So focal length is "f" and the object distance is "2f." Do a little algebra and find the image.
 
Would the focal length be equal to the object distance?
 
leehom said:
Would the focal length be equal to the object distance?

No, try again. You're close.

But I have to go now, keep at it.
 
Okay, thnx for your help
 
All you have to do is substitute 2f = do into the equation you listed above and solve for di.
 
Last edited:

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