Where is the image formed and what is its magnification?

Click For Summary
SUMMARY

The discussion centers on the formation of an image using a convex lens with a focal length of 20cm and an object placed 10cm in front of it. The lens formula, 1/f = 1/u + 1/v, is applied to determine that the image distance (v) is -20cm, indicating the image is virtual and located 20cm on the same side as the object. The magnification is calculated as 2, confirming the image is inverted. Participants emphasize the importance of understanding the lens equation and the significance of the negative sign in the image distance.

PREREQUISITES
  • Understanding of the lens formula: 1/f = 1/u + 1/v
  • Knowledge of convex lens properties and image formation
  • Familiarity with magnification calculations: m = -di/do
  • Ability to draw ray diagrams for lens systems
NEXT STEPS
  • Study the derivation and applications of the lens formula in optics
  • Learn how to construct accurate ray diagrams for various lens configurations
  • Explore the implications of positive and negative magnification in image formation
  • Investigate the differences between real and virtual images in optical systems
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding lens behavior and image formation principles.

suf7
Messages
66
Reaction score
0
Need Help With Lens!

An object is placed 10cm in front of a convex lens of focal length 20cm. Where is the image formed and what is its magnification??..Your solution should include an accurately drawn ray diagram?

Pleeez help me with this question?...im so stuck??

Thanks.
 
Physics news on Phys.org
Use the lens formula:

1/f = 1/u + 1/v

where f = focus, u = object distance from lens, and v = image distance from lens

So:

1/20 = 1/10 + 1/v

Solve for v
v = -20cm

So the image is located at 20cm
 
Last edited:
suf7 said:
An object is placed 10cm in front of a convex lens of focal length 20cm. Where is the image formed and what is its magnification??..Your solution should include an accurately drawn ray diagram?

Pleeez help me with this question?...im so stuck??

Thanks.

You should have at least[/b] be able to know where to start! The lens equation would have been a good one. If you didn't even know this, then you should review the material again and know what that equation is for, and how to use it.

If you do that, then you would have been able to derive what IntuitioN did, AND, be able to decipher what the "v = -20 cm" answer means (i.e. what does the NEGATIVE sign signifies). So look at the lens equation, and understand how each symbol are defined.

Zz.
 
IntuitioN said:
Use the lens formula:

1/f = 1/u + 1/v

where f = focus, u = object distance from lens, and v = image distance from lens

So:

1/20 = 1/10 + 1/v

Solve for v
v = -20cm

So the image is located at 20cm

oh right, thank you...so when i work out the magnification would it be 20/10 which is 2?
 
And for your ray diagram draw the lens with a compass. Then put in the object and draw the rays.
 
the ray diagram should have given you a clue to the image's location, magnification, and orientation. it is actually pretty easy.
 
magnification formula is m = - \frac{d_{i}}{d_{o}}
where di is the image distance and do is the object distance and the negative sign means that if it was positive then the image is in the same direction as the object, if neaitve then the opposite direction.
thi DIRECTION means that if the object were an arrow poiting upward, then a negatie magniification would mean that the image arrow would point downwwward, signifying an inverted image
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
10K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K
Replies
3
Views
2K