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

In summary, 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. The distance between the image and the diverging lens has been calculated to be -32cm, and it is correct. 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. The image height is half of the first image. The second image is virtual and forms in front of the second lens. The lens formula for this problem is 1/image distance - 1/object
  • #1
Incognitopad
11
0

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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  • #2
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
 
  • #3
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.
 
  • #4
Here ...
 

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  • #5
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
 
  • #6
Image distancecis coming -150/8 cm

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

V = image distance
U = object distance
F = focal length
 
  • #7
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:
  • #8
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
 

Attachments

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  • #9
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
 

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

What is the formula for calculating image height?

The formula for calculating image height is: image height = object height x magnification.

What is the difference between object height and image height?

Object height refers to the actual size of the object being viewed, while image height refers to the size of the object's image produced by the optical system.

Can image height be negative?

No, image height cannot be negative. It represents the distance from the top to the bottom of the image, and therefore must always be a positive value.

How does the distance between the object and the lens affect image height?

The distance between the object and the lens, also known as the object distance, is a factor in determining the magnification of the image. As the object distance increases, the image height decreases, resulting in a smaller image.

Is there a maximum or minimum value for image height?

There is no maximum or minimum value for image height, as it can vary depending on the object distance, lens properties, and magnification. However, it is always a positive value and cannot be zero.

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