How Can You Determine the Correct Sign for Image Height in Mirror Equations?

In summary: Ok, thanks so far, but how, in general, do I know if the height is going to be positive or general, because I keep ending up with a negative number, but in most of my questions, the answer is a positive number.It's a matter of interpreting the question. As a beginning, I would say that the variable h_i can be positive or negative, and we call that variable the height. So when they ask for the height, they are probably are asking for the value with the correct sign h_i. When they ask for the size, length, dimension, etc. I would think they probably mean just the magnitude |h_i|. But you still have to read and interpret the question
  • #1
ital_dj
31
0
[SOLVED] General question about mirrors

I'll give you a question as an example. My question to you isn't about solving the question, as you'll see below.

Homework Statement


A thumb of height 8cm is held in front of a concave mirror of focal length 10cm. The image formed is 12cm from the vertex of the mirror. Find:
[I'll give you the one that applies to the question]
c) the size of the image

Homework Equations


[tex]\frac{hi}{ho}[/tex] = [tex]\frac{-di}{do}[/tex]

The Attempt at a Solution


I plugged in 8cm under ho, 12cm under di, and 60cm under do (calculated previously), and came out with the answer -1.6cm. The answer in the book says it's positive 1.6cm. I've made this mistake previously, and was wondering how I can differentiate between the positive and negative solution.

Thanks in advance.
 
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  • #2
well either your di or your do should be negative because they are on opposite sides of the mirror. its up to you how you make your axes. at least that's what it seems like. id recommend reading the text where you got the equation to see how they define di and do (whether its a distance or displacement).
 
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  • #3
genghiskron said:
well either your di or your do should be negative because they are on opposite sides of the mirror. its up to you how you make your axes. at least that's what it seems like. id recommend reading the text where you got the equation to see how they define di and do (whether its a distance or displacement).

But then when finding something like magnification, if the di or do is a negative, then the magnification (formula M= -di/do ) would be a positive number, which ends up being wrong because the image is supposed to be inverted, but ends up erect.

EDIT: di and do is defined as distance
 
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  • #4
Hi ital dj,

It might help if you could you post the rest of the problem. With only what's given it leaves a few questions. Like how do you know the image is 12 cm in front of the mirror instead of 12 cm behind?

More importantly, if part c asked for the size and part d asked for the orientation (erect/inverted), then that would be a good indicator that in part c they were using the word size to mean the magnitude of the height [itex]h_i[/itex].
 
  • #5
alphysicist said:
Hi ital dj,

It might help if you could you post the rest of the problem. With only what's given it leaves a few questions. Like how do you know the image is 12 cm in front of the mirror instead of 12 cm behind?

More importantly, if part c asked for the size and part d asked for the orientation (erect/inverted), then that would be a good indicator that in part c they were using the word size to mean the magnitude of the height [itex]h_i[/itex].

I gave you the whole question, the only things I left out were: a) the position of the object; b) what sign should the focal length have; and d) the type and orientation of the image
 
  • #6
The question in part d is what I was looking for. When you solve for the height [itex]h_i[/itex], it gives you the size (magnitude of height) and the orientation (from the sign of h).

In this problem it looks like they split these parts up: they just want the size for part c, and they ask about the orientation (which is just asking about the sign of h) in part d.
 
  • #7
alphysicist said:
The question in part d is what I was looking for. When you solve for the height [itex]h_i[/itex], it gives you the size (magnitude of height) and the orientation (from the sign of h).

In this problem it looks like they split these parts up: they just want the size for part c, and they ask about the orientation (which is just asking about the sign of h) in part d.

Ok, thanks so far, but how, in general, do I know if the height is going to be positive or general, because I keep ending up with a negative number, but in most of my questions, the answer is a positive number.
 
  • #8
I don't believe there are any hard rules; it's a matter of interpreting the question. As a beginning, I would say that the variable [itex]h_i[/itex] can be positive or negative, and we call that variable the height. So when they ask for the height, they are probably are asking for the value with the correct sign [itex]h_i[/itex]. When they ask for the size, length, dimension, etc. I would think they probably mean just the magnitude [itex]|h_i|[/itex].

But you still have to read and interpret the question carefully to be sure for a specific case.
 

FAQ: How Can You Determine the Correct Sign for Image Height in Mirror Equations?

What are mirrors made of?

Mirrors are typically made of a thin layer of metal, such as aluminum or silver, that is deposited on the back of a glass surface. This metal layer reflects light, creating the image we see.

Why do mirrors reflect images?

Mirrors reflect images because of the law of reflection, which states that when light hits a smooth, flat surface, it bounces off at the same angle it hit the surface. This creates a clear, reversed image of the object in front of the mirror.

Do all mirrors reflect light equally?

No, not all mirrors reflect light equally. The type of metal used, the thickness of the glass, and the quality of the surface all play a role in how much light is reflected. Mirrors used in telescopes, for example, are designed to reflect as much light as possible.

Can mirrors be used to create an infinite reflection?

Yes, mirrors can be used to create an infinite reflection. This is known as the "mirror effect" and is achieved by placing two mirrors facing each other at a slight angle. The reflected images will continue to bounce back and forth, creating the illusion of an infinite number of reflections.

Can mirrors be used for purposes other than reflection?

Yes, mirrors can be used for purposes other than reflection. They are also used in various technologies, such as lasers and solar panels, where they are used to direct and focus light. Mirrors can also be used for decorative purposes and in art installations.

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