Product of magnifications = 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.

 

[Nickie]

As a scientist, it is important to understand the reasoning behind mathematical equations and not just blindly apply them. In this case, the thin lens equation and the magnification equation have specific meanings and implications.

The thin lens equation, 1/f = 1/do + 1/di, relates the focal length of a lens (f) to the distance of the object (do) and the distance of the image (di). This equation can be used to determine the location of the image formed by a lens. However, it is important to note that this equation assumes a thin lens, which means that the thickness of the lens is negligible compared to its focal length. If the lens is not thin, this equation may not accurately predict the location of the image.

The magnification equation, m = -di/do, relates the size of the image (di) to the size of the object (do). A negative magnification indicates an inverted image, while a positive magnification indicates an upright image. However, this equation assumes that the image is formed at the same distance as the focal length of the lens (di = f). If the image is formed at a different distance, the magnification may not accurately reflect the size of the image.

In this question, the product of the magnifications is expected to be close to 1 because of the assumptions made in the thin lens equation and the magnification equation. If the lens is thin and the image is formed at the focal length, then the magnification will be equal to -1 (inverted image with equal size). Multiplying two magnifications of -1 will give a product of 1. This is a coincidence and does not necessarily mean that the product of magnifications will always be equal to 1.

It is also important to note that in real-world situations, lenses may not always produce perfect images due to factors such as lens imperfections, aberrations, and diffraction. These factors can affect the location and size of the image, and may result in a product of magnifications that is not equal to 1. As a scientist, it is important to understand the limitations and assumptions of mathematical equations and to always critically evaluate the results obtained.