Calculating Magnification with lens, an object, and a screen.

Click For Summary
SUMMARY

The discussion focuses on calculating magnification using a lens with a focal length of 9 cm, where an object is placed at a distance of 3.7 times the focal length (d = 3.7fo) from the lens. The initial magnification is determined using the thin lens equation, 1/f = 1/s + 1/s', and the magnification formula, m = -(s/s'). Participants clarify that moving the lens causes the object and image distances to switch, which is crucial for solving the problem. The discussion emphasizes understanding the relationship between object distance, image distance, and lens movement.

PREREQUISITES
  • Understanding of the thin lens equation (1/f = 1/s + 1/s')
  • Knowledge of magnification formula (m = -(s/s'))
  • Familiarity with concepts of object distance and image distance
  • Basic principles of optics and lens behavior
NEXT STEPS
  • Study the derivation and applications of the thin lens equation
  • Explore the concept of magnification in different optical systems
  • Investigate the effects of lens movement on image formation
  • Learn about practical applications of lenses in photography and microscopy
USEFUL FOR

Students studying optics, physics enthusiasts, and anyone looking to deepen their understanding of lens behavior and image formation principles.

jmacmartin
Messages
4
Reaction score
0

Homework Statement



A lens with focal length fo = 9 cm forms an image of an object placed a distance d=3.7fo to the left of the lens, onto a screen to the right of the lens. The distance L from object to image is fixed. The lens is now moved toward the screen a distance x until an image is formed again. (a) What is the magnification in the initial case? (b) How far was the lens moved? (c) What is the magnification in the second case? (d) What is the distance L between the object and the image?

Homework Equations



1/f = 1/s + 1/s'

and

m = -(s/s')

The Attempt at a Solution



I just figured out part A and am now stuck on part B. I'm not sure how to go about solving it. I read online that "when the lens is moved, the object and image distances are switched." Can someone please explain to me why this is? I can't seem to find any information in my book. Not looking for the answer, but more how to understand what happens as this lens is moved.

Thanks!
 
Last edited:
Physics news on Phys.org
It comes from the thin lens equation:

1/f = 1/s + 1/s'

You can switch the object and image distances and the equation still holds.
So moving the lens until the image distance is equal to the previous object
distance will result in a new object distance equal to the previous image distance.
 

Similar threads

Focal length of the eyepiece lens in a compound microscope
  • · Replies 7 ·
Replies
7
Views
2K
Find the location and length of the final image
  • · Replies 2 ·
Replies
2
Views
1K
Analyzing this 2-lens optical system
  • · Replies 5 ·
Replies
5
Views
2K
Calculate a radius of a circle on a screen
  • · Replies 3 ·
Replies
3
Views
1K
Mirror placed horizontally between two half convex lenses
  • · Replies 11 ·
Replies
11
Views
2K
Optics - Adding magnification in an optical system
  • · Replies 3 ·
Replies
3
Views
2K
Calculate the focal length of a thin convex lens
  • · Replies 19 ·
Replies
19
Views
1K
How to measure the magnification of a microscope in the lab
  • · Replies 12 ·
Replies
12
Views
3K
Inserting thick lenses into a thin lens system and deducing values
  • · Replies 1 ·
Replies
1
Views
2K
Calculating Magnification: Diverging and Converging Lenses Equations Explained
  • · Replies 3 ·
Replies
3
Views
2K