# Lateral Magnification

1. Aug 10, 2008

### Gear2d

1. The problem statement, all variables and given/known data

If lateral magnification is negative what does that mean?

2. Relevant equations

m = -di/do

3. The attempt at a solution

From what I understand that focal distance for converging devices (concave mirrors and convex lens) is always positive. So this would make m="-" since di= "+". Also that a negative lateral magnification would mean less power. My question are:

1) Is the above correct?

2) If the object was within the focal point the image would be virtual and upright, what would happen to power then?

3) If I increase the focal distance would that mean I would decrease the power of the lens or mirror based on this formula: P=1/f?

4) Based on this formula: P=1/f = 1/di + 1/do, the farther the object is from the lens or mirror the greater power is?

2. Aug 10, 2008

### alphysicist

For a single lens or mirror: When $d_i$ is positive, then $m$ would be negative; but $d_i$ can be negative, in which case $m$ would turn out to be a positive number.

If you have a system of lenses or mirrors, then $d_o$ can also be negative.

The power of a lens is a property of that lens; it's the inverse of the focal length (having units of diopters when focal length is in meters). So the power of a particular lens is a constant.

Just to repeat, the power would be the same.

However, go ahead and try out this case. Set $f=10$ and $d_o$ be anything less than 10, for example. Solve for $d_i$, and then $m$. What is the sign of $m$?

Do the same for $f=10$ and do be anything greater than 10, and find the sign of $m$.

So what does the sign of $m$ mean?

Okay, but to change the focal length you have to change the radius of curvature of the sides of the lenses; so effectively you have a different lens altogether.

Do you now see why this is not true?

3. Aug 10, 2008

### Gear2d

Thanks alphysicist. I see what you are saying.