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Magnification/Focal Length

  • Thread starter matt72lsu
  • Start date
  • #1
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Homework Statement



Shaving/makeup mirrors typically have one flat and one concave (magnifying) surface. You find that you can project a magnified image of a lightbulb onto the wall of your bathroom if you hold the mirror 2.0 m from the bulb and 3.7 m from the wall.

a) What is the magnification of the image?
b) Is the image erect or inverted?
c) What is the focal length of the mirror?

Homework Equations



m = - di/do ???
1/di +1/di = 1/f

The Attempt at a Solution


I'll be honest, I have no clue how to start this problem. Any help would be appreciated
 

Answers and Replies

  • #2
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  • #3
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Ok thanks! as far as magnification, is the formula i provided correct? and how do i figure out if it is inverted or not?
 
  • #4
648
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The positive or negative signs in the equations have the following significance

(from further down in the link I gave you)

• f is + if the mirror is a concave mirror
• f is - if the mirror is a convex mirror
• di is + if the image is a real image and located on the object's side of the mirror.
• di is - if the image is a virtual image and located behind the mirror.
• hi is + if the image is an upright image (and therefore, also virtual)
• hi is - if the image an inverted image (and therefore, also real)


hi is height of image, ho is height of object

The formula for these is
magnification = hi/ho = -di/do

Just keep a close eye on the negative sign in the equation and be consistent with signs generally, and this will tell you if the image is real and inverted; or virtual and erect.

The other way of working this out is to draw a ray diagram.
 

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