Radius of convex spherical mirror?

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Homework Help Overview

The problem involves determining the radius of a convex spherical mirror based on the distance from the mirror to an object and the height of the image required for identification. The context includes measurements in feet and inches, with a focus on the properties of convex mirrors.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the application of mirror formulas, including the mirror equation and magnification. There are attempts to clarify the signs associated with image distance and height, particularly regarding the nature of the image produced by a convex mirror.

Discussion Status

The discussion has explored various interpretations of the mirror's properties, with participants questioning the correctness of their calculations and assumptions about image orientation. Some guidance has been provided regarding the characteristics of virtual images produced by convex mirrors.

Contextual Notes

There is confusion regarding the signs used in calculations, particularly in relation to the nature of the image (virtual vs. real) and the implications for the focal length and radius. Participants are also considering the practicality of the calculated sizes for a convenience store mirror.

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The problem:
"A convex spherical mirror is 25 ft from the door of a convenience store. The clerk needs to see a 6 ft. person entering the store at least 3 inches tall in the mirror to identify them. What is the radius of the mirror?"

d_obj = do = 25 ft = 300 inches
h_img = hi = 3 inches
h_obj = ho = 6 ft = 72 inches

The formulas I tried using:

focal point = f = radius/2 = r/2
1/do+1/di = 1/f = 1/r/2 = 2/r
For convex mirror: m = magnification = hi/ho = -di/do

Work:

hi/ho = -di/do
3/72 = -di/300
di=-12.5 inches

1/do + 1/di = 2/r
1/300 - 1/12.5 = 2/r
r=26 inches

I found the radius to be 26 inches, which seems a bit too large. Is it correct or am I doing something wrong? I feel as though I must somehow incorporate the magnification, but I'm not sure how. It would be -24X but looking at something normally you'll see it as smaller than it is if you're at a greater distance so I'm not sure how to compensate for that.
 
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The image is real and upside down, so the height of the image is negative. Therefore, di is positive.

ehild

Edit: I was wrong: The mirror is convex, so the image is upright (virtual). Di is positive!
 
Last edited:
ehild said:
The image is real and upside down, so the height of the image is negative. Therefore, di is positive.

ehild

d_obj = do = 25 ft = 300 inches
h_img = hi = -3 inches
h_obj = ho = 6 ft = 72 inches

hi/ho = -di/do
-3/72 = -di/300
di=12.5 inches

1/do + 1/di = 2/r
1/300 + 1/12.5 = 2/r
r=24 inches

So the radius of the mirror is precisely 24 inches? That still seems like a rather large convenience store mirror.
 
It is the radius of the curvature. A spherical mirror is a part of a spherical surface, its diameter can be as small as you like.

ehild
 
Okay, thank you!
 
Sorry I was wrong, mixed convex and concave. The clerk does not want to see people upside down! He/she uses a convex mirror, that produces virtual, upright image. For such an image, di is positive, but the focal distance is negative. From the mirror equation, you get f=-13 inch. So the 26 inch for the radius was correct.

ehild
 
Okay, thanks. I was wondering about the inverted image, though the result seemed better.
 

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