Solve for Image Location and Magnification in Concave Mirror | Homework Help

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SUMMARY

The discussion focuses on solving for the image location and magnification of an object placed in front of a concave mirror with a radius of 12.0 cm. The mirror equation, 1/f = 1/do + 1/di, is applied, leading to a focal length (f) of 6 cm and an image distance (di) calculated as -12 cm. The magnification (m) is determined using the formula m = hi/ho = -di/do, where the object distance (do) is 4.0 cm. Participants emphasize the importance of drawing a ray diagram to visualize the image formation process.

PREREQUISITES
  • Understanding of the mirror equation (1/f = 1/do + 1/di)
  • Knowledge of magnification formula (m = hi/ho = -di/do)
  • Familiarity with ray diagrams for concave mirrors
  • Basic concepts of focal length and image distance
NEXT STEPS
  • Study the construction and interpretation of ray diagrams for concave mirrors
  • Explore the effects of varying object distances on image characteristics in concave mirrors
  • Learn about the applications of concave mirrors in real-world scenarios
  • Investigate the relationship between object height and image height in mirror systems
USEFUL FOR

Students studying optics, physics educators, and anyone seeking to understand image formation in concave mirrors.

Jonah14
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Homework Statement



"An object is 4.0 cm in front of a concave mirror having a 12.0-cm radius. Locate the image using the mirror equation and a ray diagram."

Homework Equations



1/f = 1/do + 1/di where f = Focal Length, do = Distance Object, di = Distance Image

m = hi/ho = -di/do where m = Magnification, hi = Height of Image, ho = Height of Object

The Attempt at a Solution



f = 6 cm
do = 4.0 cm
Ho = ?
di = -12
Hi = ?
C = 12



I got 1/6 = 1/di + 1/4
1/6 - 1/4 = 1/di
Which equaled -12


Hi/Ho = -(-12)/4
Usually I knew Hi or Ho to get the answer but I don't know it this time!
Can anyone help me with this seemingly easy problem?
 
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First of all, you found the answer, you just need d this time and show where it is.

Now then also probably want a diagram there, just to test your understanding of the geometry of the problem.

Ok, first of all you need to find the image, so you need to draw your ray diagram.

I wish I could post an image for this, but physicsforums won't let me (you can only link to one), which is a little lame.

But if you draw a concave mirror, then the focal point, where will be the image compared to it?

Then draw the image as a perpendicular arrow going up (your book must show this somewhere). Now you just need to draw three rays:
1. Draw a ray going straight to the mirror from the top of the arrow you drew (perpendicular to the arrow you drew as well). The rule of this ray is that it will bounce to the focal point, so when it hits the mirror, draw a straight line from there to the focal point


2. Draw a ray that goes from the top of the arrow through the focal point of the mirrror. The rule for this ray is that once it hits the mirror, it bounces back parrallel to the ground.

3. Draw a ray that goes from the top of the arrow you drew again, but this time going to the center of the mirror. The rule for this one is that you draw a line coming back at the same angle is came in, but downwards


With these three lines you have, elongate them until they intersect. They may intersect in front of or behind the mirror, so you must elongate them both ways. Where they intersect is the top part of the new arrow you will draw perpendicular to the ground, your image.

Aah, I'll just draw an example which will be simpler. Use this as a guideline to solve yours.
NeYPv.jpg


And now, for your problem, where is the image located, is it contestant #1, #2, #3?

JgQog.jpg
 

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