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Confused on mirror/lens sign conventions

  • Thread starter Amith2006
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427
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Sir,
I am confused on what sign conventions to follow. I reffered 5 or six books which says that focal length of convex mirror is positive and focal length of concave mirror is negative. But in Resinick & halliday,wikipedia and many other websites it is given the opposite. Please help me!!!!!!!!!
 

Hootenanny

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I've always used the convention described in Resinick & halliday.

  1. A convex mirror has a negative focul legnth.
  2. A concave mirror has a positive focul length.
 
427
2
Sir,
In the previous thread I had asked you to see if the sign conventions were right and you said it was right. But in that I had used wrong sign conventions.
 

Hootenanny

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Are you refering to this question?

Amith2006 said:
2)A convex mirror of focal length f produces an image (1/n)th of the size of the object. What is the distance of the object from the mirror?
I solved it in the following way:
Magnification = 1/n = -(v/u)
i.e. v = -(u/n)
1/u + 1/v = 1/f
1/u – (n/u) = 1/f
By solving I get,
u = (1-n)f
Are my sign conventions right? Sometimes the diagram may not be clear. So I will try to describe the diagram. A convergent beam of light serves as a virtual object which appears to converge at a distance of 10 cm behind the mirror.
In which case I only checked you manipulation (I thought that was what you wished to be checked). You didn't put any values in for 'f' or 'o' etc, therefore it is not possible to say whether you have used correct sign convention.
 
427
2
Sir,
Then I am sorry Sir. Now I have a doubt. It is said that a convex mirror always forms virtual images of a real object. But what about the nature of the images formed by it of a virtual object? Is it always real?
 

Hootenanny

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A virtual object forms a real image if it is inside the focul length of the mirror.
 
427
2
Sir,
Is this applicable to both convex and concave mirrors?
 

Doc Al

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Amith2006 said:
Is this applicable to both convex and concave mirrors?
You should be able to answer this for yourself by examining the mirror equation:
[tex]\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}[/tex]
or:
[tex]\frac{1}{d_i} = \frac{1}{f} - \frac{1}{d_o} [/tex]

For a virtual object, [itex]d_o < 0[/itex]. Ask yourself for what object distances do the image distances become positive, thus forming real images. You'll find that convex mirrors ([itex]f < 0[/itex]) behave differently than concave mirrors ([itex]f > 0[/itex]).

(Note that I use the same sign convention that Hootenanny described.)
 
Last edited:
427
2
Thank you very much Sir.
 
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The best book I've read from which completely clarifies any doubts about sign conventions in geometric optics is Resnick+Halliday+Krane Volume 2. If you are from India, you most likely use the old cartesian convention (which for some crazy reason is called the new cartesian convention). This is prone to errors and misinterpretation. The key point here is that the results should be independent of the convention. Yet the convention is just as important as the equations are.

I always use the following convention for lenses:

If a refracting surface is concave towards incident light (opens towards the incident light) then its radius of curvature is negative. This is true for a concave refracting surface.

If a refracting surface is convex towards incident light (opens away from the incident light) then its radius of curvature is positive. This is true for a convex refracting surface.

Note that I have said nothing about the focal length which depends on the refractive indices of the media on either side of the surface. So I cannot say anything about the focal length of a lens offhand either.

Hope this helps....
 

Doc Al

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maverick280857 said:
Note that I have said nothing about the focal length which depends on the refractive indices of the media on either side of the surface. So I cannot say anything about the focal length of a lens offhand either.
maverick is right that the focal length of a lens depends on the relative indices of refraction of the lens and the surrounding media. (For details, study the Lens-Maker's formula: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenmak.html or read Halliday & Resnick.)

But for simple situations--what I assume we are discussing here--where the lens has a higher refractive index than the surrounding media (like a glass lens in air), you can certainly deduce that:
a convex lens converges parallel rays and thus has a positive focal length
a concave lens diverges parallel rays and thus has a negative focal length​

The key point to remember is to follow the light. If the lens focuses a beam of parallel light on the side where the light actually goes, then its focal length is positive. (Imagine parallel light coming in from the left, passing through the lens, then continuing to the right. If the focal point is on the right of the lens, then it's a converging lens with positive focal length.)

Same with a mirror, except that the light bounces off a mirror. Again, imagine parallel light coming in from the left, hitting the mirror, then reflecting back to the left. If the focus is on the left (where the light actually goes), then the focal length is positive (it's a converging mirror).
 

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