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Here is the question as given in my textbook:

Find the distance of the object from a concave mirror of focal length 10 cm so that the image size is 4 times the size of the object.

The solution in my textbook has the following data stated:

$u=-x$ as it is assumed that the object is real.

$v=-4x$ as it is assumed in case 1 that a real image will be formed and $|\frac{v}{u}|=|m|=4$

So now I am not able to understand why the image distance $v$ is taken $-4x$. In the question it is given that the object size is magnified 4 times, and not the distance of the of the object from the mirror.

And the magnification formula is $\frac{-v}{u}=m$, so why does the solution include the modulus of the formula? Am I missing something in analysing the solution?

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  • $\begingroup$ Unfortunately, many solution sets not written with the intention of being a primary teaching tool. People who already know really well how to solve this problem know what to do with the solution, but the notation and the negative signs and absolute values are likely different from the notation in your textbook and in class if I had to guess... $\endgroup$ – jdj081 Oct 8 '13 at 5:09
  • $\begingroup$ this is the first solution on the following topic and i know this topic very well but i am really getting confused on this following solution i think i am not able to link something in the solution $\endgroup$ – Dimensionless Oct 8 '13 at 7:36
  • $\begingroup$ Strongly recommend that you draw a diagram; the issue of signs is very well explained in @Gregsan's answer, and together with diagram should lead to solution/understanding. $\endgroup$ – Floris Jul 15 '14 at 12:10
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$u$ is the distance of the object to the reflecting/refracting surface. $v$ is the distance from that surface to the image.

by convention

  1. object and image: if both are real, both must have the same sign. if one is real and the other virtual, they must have different sign. for mirrors, both being on the same side of the mirror means they have the same property. for lenses, both being on the same side means they have different properties.
  2. concave mirrors have positive $f$ and convex mirrors have negative $f$.
  3. for magnification, positive values indicate that the image is the same orientation as the object (object upright, image also upright), negative if the image is inverted relative to the object. this use of sign is analogous to signs on vectors to indicate direction on an axis. if the magnification formula has a modulus on it, it means that the information regarding uprightness or inversion is deemed unimportant for the time being, or whatever other reasons.

knowing these rules is paramount to properly understanding optics problems as they are posed.

in this question the magnifcation is given as 4, whereas information on orientation is unknown.therefore the sign of $v$ is unknown (i am assuming there was no diagram that came along with the question). since u is assigned a negative value $-x$, then $v$ must also be a negative value if the image is real (now the information is given to you). this is interpreted as both object and image being on the same side of the mirror, on the x axis. proper use of the formula $m=\frac{-v}{u}$ will yield a negative value for m, indicating an inverted image relative to the object. in fact it is implicit from the rules above that for concave mirrors, real images are always inverted--an upright image must be a virtual image.

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  • $\begingroup$ i know all that i am good at this topic but i am getting confused on the following assumption's made in text-book $ v=−4x$ as it is assumed in case 1 that a real image will be formed and $|\frac{v}{u}|=|m|=4$ $\endgroup$ – Dimensionless Oct 8 '13 at 8:03
  • $\begingroup$ please read 1. and 3. $\endgroup$ – gregsan Oct 8 '13 at 8:05
  • $\begingroup$ if there is no good reason to assume the image will be real, then two answers can be posed: v=-4x if real image or v=4x if virtual image. $\endgroup$ – gregsan Oct 8 '13 at 8:07
  • $\begingroup$ Great example of how to answer homework question: help in understanding without feeding the solution. $\endgroup$ – Floris Jul 15 '14 at 12:11
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in concave lens magnification cannot be negative because if it is negative the image should be inverted and real as in the convex lens but it is not true in concave lens. As the concave lens the image will be always virtual and erect if magnification is positive.

if the sign changes the lens cannot become a convex lens, what is true in mathematics cannot be true in giving sufficient inference.

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$u=-x$ ,here $x$ is the value of object distance, as per sign convention $x$ becomes $-x$ because $x$ is being left of the mirror. $v=-4x$, it came form $m=-\frac{v}{u}$, $4=-v/x$, (We know that $m=4$ and $u=-x$ here $u=x$ because this is formula is already sign covensed so no need to change $x$ to $-x$) $4x=-v$, then comes $-4x=v$

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