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

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In the discussion about the image height for a converging and diverging lens system, a converging lens with a focal length of 30 cm and a diverging lens with a focal length of -64 cm are positioned 139 cm apart, with a 4.0 cm tall object placed 50 cm in front of the converging lens. The calculated image distance from the diverging lens is -32 cm, leading to a height of 2.56 cm for the image formed. However, the correct answer is stated to be 3.0 cm, prompting confusion about the calculations. Participants clarify that the magnifications from both lenses must be considered, and the second image formed is virtual, requiring ray diagrams to accurately determine its position. The discussion emphasizes the importance of understanding lens conventions and the proper application of formulas in optics.
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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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