What is the Image Height for a Converging Lens and Diverging Lens System?

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Homework Help Overview

The discussion revolves around a problem involving a converging lens and a diverging lens system, where participants are trying to determine the image height produced by the system. The setup includes specific focal lengths and distances between the lenses and the object.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants share their calculations for image height and discuss the role of the diverging lens in the system. Some express confusion about the image distances and magnifications involved.

Discussion Status

There are multiple calculations presented, with some participants arriving at different image heights. Guidance has been offered regarding the need to consider the diverging lens and the nature of virtual images. Participants are exploring various interpretations of the problem.

Contextual Notes

Some participants question the assumptions regarding the ray paths and the necessity of the diverging lens in the calculations. There is also mention of differing conventions in drawing ray diagrams and defining magnification.

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Homework Statement



A converging lens with a focal length of 30cm and a diverging lens with a focal length of -64cm are 139cm apart. A 4.0cm tall object is 50cm in front of the converging lens.

distance between image and diverging lens has been calculated to be -32cm, and it is right.


Homework Equations



M = -(s'/s) = (hi/ho)

The Attempt at a Solution



-(-32cm/50cm) = (hi/4.0cm)
hi = 2.56cm

I already got the mastering physics problem wrong. Used all my guesses. But the answer is 3.0cm and I have no clue how?
 
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Incognitopad said:

Homework Statement



A converging lens with a focal length of 30cm and a diverging lens with a focal length of -64cm are 139cm apart. A 4.0cm tall object is 50cm in front of the converging lens.

distance between image and diverging lens has been calculated to be -32cm, and it is right.


Homework Equations



M = -(s'/s) = (hi/ho)

The Attempt at a Solution



-(-32cm/50cm) = (hi/4.0cm)
hi = 2.56cm

I already got the mastering physics problem wrong. Used all my guesses. But the answer is 3.0cm and I have no clue how?

The magnifications multiply, and it is not the same as the image distance from the second lens divided by the object distance from the first lens.

ehild
 
I also calculated it and it came out to be 2.5 cm ... from what i see there is no role of concave lens. I took it as a gimmik becouse the image would never reach the concave lens . Its a gimmik.
 
Here ...
 

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silvercrow said:
I also calculated it and it came out to be 2.5 cm ... from what i see there is no role of concave lens. I took it as a gimmik becouse the image would never reach the concave lens . Its a gimmik.

You did something wrong.
The first image is at distance of 75 cm after the converging lens. The magnification is N1=-di/do=-75/50. The image height is 6 cm and it is inverted. The first image is object for the diverging lens.
The object distance from the diverging lens is 139-75=64 cm. The second image distance is -32, as you got. N2=-di/do=32/64. The image height is half of the first image. ehild
 
Image distancecis coming -150/8 cm

By the formula 1/v - 1/u = 1/f

V = image distance
U = object distance
F = focal length
 
Wait i was taking focal length as negative :P . I started this chappter two days ago ... but the ray is not reaching the concave lens so why consider it ?
 
Last edited:
silvercrow said:
Wait i was taking focal length as negative :P . I started this chappter two days ago ... but the ray is not reaching the concave lens so why consider it ?

The ray reaches the second lens if it is big enough. Or draw those rays which reach it :)
The focal length of the converging lense is positive, that of the diverging lens is negative.
The second image is virtual: It forms in front of the second lens. You have to draw the rays backwards to get the intersection.

ehild
 

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ehild said:
The ray reaches the second lens if it is big enough. Or draw those rays which reach it :)
The focal length of the converging lense is positive, that of the diverging lens is negative.
The second image is virtual: It forms in front of the second lens. You have to draw the rays backwards to get the intersection.

ehild

Your ray diagram seems tovbe wrong because according to convention rays are drawn left to right :)
 
  • #10
silvercrow said:
Your ray diagram seems tovbe wrong because according to convention rays are drawn left to right :)

He-he. In my country,they can go in any direction, (the arrows show the direction); moreover we use the lens formula 1/t+1/k =1/f and define magnification as N=k/t and it is positive when the image is inverted.:biggrin:

Have you understood that second image?

ehild
 
  • #11
ehild said:
He-he. In my country,they can go in any direction, (the arrows show the direction); moreover we use the lens formula 1/t+1/k =1/f and define magnification as N=k/t and it is positive when the image is inverted.:biggrin:

Have you understood that second image?

ehild

What is t and k ?
 
  • #12
The same as object distance and image distance.

ehild
 
  • #13
No formula for lens is 1/image distance - 1/ object distance = 1/f
 

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