# Calculate the image height

## 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?

ehild
Homework Helper

## 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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ehild
Homework Helper
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:
ehild
Homework Helper
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

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 :)

ehild
Homework Helper
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. Have you understood that second image?

ehild

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. Have you understood that second image?

ehild

What is t and k ?

ehild
Homework Helper
The same as object distance and image distance.

ehild

No formula for lens is 1/image distance - 1/ object distance = 1/f