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Image Distance and Magnification

  • #1
Question

An object is a distance of 23.4 cm from the center of a silvered spherical glass Christmas tree ornament which has a diameter of 5.70 cm.

A What is the position of its image?

B What is the magnification of its image?

I think the object can be treated as a convex mirror.

s (object distance) = 23.4 cm
R = -(5.70)/2 = -2.85 cm

[tex]\frac{1}{s} + \frac{1}{s'} = \frac{2}{R}[/tex]

Using that equation, I get [tex]\frac{1}{s'} = \frac{2}{-2.85} - \frac{1}{23.4}[/tex].

Then, s' = -1.12 cm.

And, [tex]m = -\frac{s'}{s}[/tex], so m = .048.

I know for sure that -1.12 cm is not the correct answer, so most likely my second answer is also incorrect.

What did I do wrong?
 

Answers and Replies

  • #2
Did I not provide enough work?
 
  • #3
Doc Al
Mentor
44,939
1,200
Your method is fine but check your arithmetic.
 
  • #4
Sorry, I still get the same answer.
 
  • #5
Doc Al
Mentor
44,939
1,200
Using that equation, I get [tex]\frac{1}{s'} = \frac{2}{-2.85} - \frac{1}{23.4}[/tex].

Then, s' = -1.12 cm.
Please redo this calculation.
 
  • #6
Doc Al,

Indeed, it was incorrect. Thanks!
 

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