Calculating Magnification: Diverging and Converging Lenses Equations Explained

I love them (that's actually what I'm studying in school).In summary, the task is to find the magnification of a system consisting of a diverging lens and a converging lens. The focal points of the lenses are -9.5 cm and 13.0 cm, respectively, and the diverging lens is placed 4.35 cm to the right of the converging lens. The equations used are M = -p/q, 1/p + 1/q = 1/f, p of diverging lens = d - q of converging lens, and Mtotal = M1 * M2. The magnification of the diverging lens is 11.17647059, and the magnification of the
  • #1
brinstar
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1

Homework Statement


"Find the magnification of the following system: A diverging lens with focal point -9.5 cm is placed 4.35 cm to the right of a converging lens with its own focal point of 13.0 cm. Parallel light enters the converging lens from the left."

Homework Equations


M = -p/q
1/p + 1/q = 1/f
p of diverging lens = d - q of converging lens
Mtotal = M1 * M2

The Attempt at a Solution


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So I figured out the magnification for the diverging lens and got 11.17647059. I cannot for the life of me figure out the magnification of the converging lens, and this is all because the object distance is infinite. How exactly do I get the total magnification (which is the product of my two magnifications) if the object distance for my converging lens is infinite? That would leave me at 13.0 cm divided by infinity...
 
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  • #2
Magnification (of the whole telescope together)= fo/fe

fo is the focal length of the objective lens (converging)
fe is the focal length of the eyepiece lens (diverging)

The object distance doesn't matter.
 
  • #3
DarkMatter5 said:
Magnification (of the whole telescope together)= fo/fe

fo is the focal length of the objective lens (converging)
fe is the focal length of the eyepiece lens (diverging)

The object distance doesn't matter.

Wow! >.< I feel so dumb.

Thanks so much for the reply!

For some reason, though, my teacher never gave me that formula in class. The ones I gave were all I had to work with. I didn't know you could get magnification without object distances and height!
 
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  • #4
No problem :oldbiggrin:. Now you know! Feel free to message me if you have any more telescope or astro questions.
 
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Related to Calculating Magnification: Diverging and Converging Lenses Equations Explained

What is magnification in optics?

Magnification in optics refers to the ratio of the size of an image produced by a lens or mirror to the size of the original object. It is typically represented by the letter "m" and can be either positive or negative, depending on whether the image is upright or inverted.

How do you calculate the magnification of a diverging lens?

The magnification of a diverging lens can be calculated using the equation m = -di/do, where di is the image distance and do is the object distance. The negative sign indicates that the image produced by a diverging lens is always smaller than the original object.

What is the equation for calculating the magnification of a converging lens?

The magnification of a converging lens can be calculated using the equation m = hi/ho = -di/do, where hi is the image height and ho is the object height. The negative sign indicates that the image produced by a converging lens is inverted.

How do you determine the focal length of a lens using the magnification equation?

To determine the focal length of a lens using the magnification equation, you need to know the object distance, image distance, and magnification. Rearranging the equation m = -di/do gives the focal length, f = -di/m. This equation can be used for both converging and diverging lenses.

Can the magnification of a lens be greater than 1?

Yes, the magnification of a lens can be greater than 1. This means that the image produced by the lens is larger than the original object. A magnification greater than 1 is typically seen with converging lenses, while a magnification less than 1 is seen with diverging lenses.

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