Why Should the Product of Magnifications Be Equal to 1?

[miaou5]

Homework Statement

There are two locations for the lens along the optical bench that will focus an image on the screen. Find one of these locations. Once you have the image in sharp focus, take measurements for: the object distance, do, the image distance, di, and the height of the image, hi.

Using the d values, calculate the magnification for each location. Place these values in the chart. Find the product of the two magnifications. NOTE: Ignore any negative signs. The value should be very close to 1. Why should the product of the magnifications be equal to 1?

Homework Equations

1/f = 1/do + 1/di

m = -di/do

The Attempt at a Solution

I've been plugging away at this question for almost an hour with no luck. I'm not sure why the product of the magnifications should equal zero...I've tried transforming the thin lens equation/doing all sorts of mathematical tricks, but I've come up with nothing. Can anybody help me with this question? :C

 

[Curious3141]

Homework Statement

There are two locations for the lens along the optical bench that will focus an image on the screen. Find one of these locations. Once you have the image in sharp focus, take measurements for: the object distance, do, the image distance, di, and the height of the image, hi.

Using the d values, calculate the magnification for each location. Place these values in the chart. Find the product of the two magnifications. NOTE: Ignore any negative signs. The value should be very close to 1. Why should the product of the magnifications be equal to 1?

Homework Equations

1/f = 1/do + 1/di

m = -di/do

The Attempt at a Solution

I've been plugging away at this question for almost an hour with no luck. I'm not sure why the product of the magnifications should equal zero...I've tried transforming the thin lens equation/doing all sorts of mathematical tricks, but I've come up with nothing. Can anybody help me with this question? :C

You have $\displaystyle \frac{1}{f} = \frac{1}{d_i} + \frac{1}{d_o}$

From the symmetry of that equation in $d_i$ and $d_o$, you should be able to see that you can swap the two variables without affecting $f$.

Hence if the first sharp image is formed when $d_i = x$ and $d_o = y$, the second sharp image will be formed when $d_i = y$ and $d_o = x$.

What is the magnification for each of those setups? Now can you see why their product is always (ideally) one?

 

[miaou5]

Yes! Thank you so much, this helps tons. A huge thank you again!

 

[Curious3141]

You're welcome.