Understanding Negative Lateral Magnification in Optics

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SUMMARY

Negative lateral magnification indicates that the image produced by a lens or mirror is inverted relative to the object. The formula for lateral magnification is m = -di/do, where di is the image distance and do is the object distance. A negative magnification suggests that the image is real and inverted, while a positive magnification indicates a virtual and upright image. The power of a lens, defined as P = 1/f, remains constant regardless of the image orientation, and increasing the focal distance results in decreased power.

PREREQUISITES
  • Understanding of optical principles, specifically focal length and magnification.
  • Familiarity with the lens maker's formula and power calculations.
  • Knowledge of real vs. virtual images in optics.
  • Basic algebra for manipulating optical equations.
NEXT STEPS
  • Explore the relationship between object distance and image distance using ray diagrams.
  • Study the effects of changing the radius of curvature on lens power.
  • Investigate the behavior of virtual images in concave mirrors and convex lenses.
  • Learn about the applications of negative magnification in optical devices.
USEFUL FOR

Students studying optics, physics educators, and optical engineers seeking to deepen their understanding of magnification and lens power relationships.

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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?
 
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Gear2d said:

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?
 
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Thanks alphysicist. I see what you are saying.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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