Understanding Negative Lateral Magnification in Optics

[Gear2d]

Homework Statement

If lateral magnification is negative what does that mean?

Homework Equations

m = -di/do

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?

 

[alphysicist]

Homework Statement

If lateral magnification is negative what does that mean?

Homework Equations

m = -di/do

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= "+".

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.

Also that a negative lateral magnification would mean less power.

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.

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?

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?

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?

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.

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?

Do you now see why this is not true?

 

[Gear2d]

Thanks alphysicist. I see what you are saying.