How Is the Image Size Calculated in a Concave Mirror?

AI Thread Summary
In the discussion about calculating image size in a concave mirror, a 23.0 mm high object is placed 31.0 cm from a mirror with a 54.0 cm radius of curvature. The mirror equation and magnification formula are used to derive the image distance and size. The image distance was found to be 209 cm, and the correct image size is determined to be 155 mm. The sign conventions indicate that a positive image distance corresponds to a real image, which is inverted, while a virtual image is upright. The absolute size of the image is calculated by taking the absolute value of the height derived from the magnification ratio.
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Homework Statement


An object 23.0 mm high is 31.0 cm away from a concave spherical mirror with a radius of curvature of 54.0 cm. What is the absolute size of the image?

Homework Equations



Maybe these are relevant

Mirror equation in terms of focal length:
\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

Magnification of image:
M=\frac{h'}{h}=-\frac{q}{p}

The Attempt at a Solution



I have already calculated the distance of the image from the mirror to be 209 cm. But how do I calculate the size of the image? Is there a special formula or do I need to use some kind of trigonometry? I had no luck using the trying the above equations...

The correct answer must be 155.
 
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Explain the notations you used: What is p and q and f? And what is a concave mirror? What about the signs of p,q and f? How is f related to the curvature of the mirror?
You know both the image distance and the object distance. How are they related to the ratio of the image height to the object hight?

ehild
 
Last edited:
ehild said:
Explain the notations you used: What is p and q and f? And what is a concave mirror? What about the signs of p,q and f? How is f related to the curvature of the mirror?
You know both the image distance and the object distance. How are they related to the ratio of the image height to the object hight?

ehild

Thank you! I got the right answer using the ratio \frac{h'}{h}=-\frac{q}{p}, but I had to exclude the minus '-' sign. Is that because the resulting image is upright (not inverted)?
 
The real image is inverted (h' is negative), the virtual image is upright (h' is positive). The image distance is positive when the image is real, and negative when the image is imaginary. You got q=209, a positive number, so h'/h was negative, h' negative, the image was real. To get the absolute size, you have to take the absolute value of h'.

ehild
 
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