Convergent lenses and calculating image position

In summary, two identical convergent lenses with a focal length of 10cm and an object height of 3.5 cm are placed 30cm apart. The distance from the object to the first lens is also 30cm. Using the equations 1/f = 1/s' +1/s and h'/h = - s'/s, the first image is determined to be 15cm in front of the first lens and has a height of -1.75cm. The second image is located 16.6cm in front of the second lens and has a height of 1.9cm. This image is real, inverted, and reduced in size. The problem also requires the use of 1/d0
  • #1
mortymoose

Homework Statement


Two convergent lens are identical in focal length (f=10cm) and the oobject height (h0) is 3.5 cm. The distance between the two lenses is 30cm and the distance from the object to the first lens is 30cm.
-Draw a diagram on the figure and show the image position (di) and size (hi) formed by the two lenses.

Homework Equations


1/f = 1/s' +1/s
h'/h = - s'/s

The Attempt at a Solution


The first place I think I am getting confused is whether or not two images are formed. When I did my work that is what i assumed.
My first step was to solve for the first image location:
1/10 = 1/30 +1/s'
s' =15cm
so, first imag eis 15cm in front of first lens
then the height of the first image:
h'/3.5cm = -15/30
h'= -1.75cm

Second image location:
1/10=1/25 +1/s'
s'= 16.6cm
the second image is 16.6cm in front of the second lens
height of second image was:
h'/-1.75cm = -16.6cm/15cm
h'= 1.9cm
So this image would be real, inverted, and reduced in size

But then I am supposed to use 1/d0 + 1/di =1/f
and that's where i realized i must be doing something wrong, because I assumed that two images would be formed but they are only asking for the one image
 
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  • #2
mortymoose said:
Second image location:
1/10=1/25 +1/s'
Check if this matches your actual problem statement. From your post #1, I conclude 15 cm, not 25.

Can you post the picture ?
 
  • #3
There is a real image formed by the lens closer to the object, then that image becomes the object for the other lens.
I believe they want you to determine the height and location of the image produced by the other lens.
 

Related to Convergent lenses and calculating image position

1. How do convergent lenses form images?

Convergent lenses are designed to bring parallel rays of light together at a single point, known as the focal point. This causes the image to appear magnified and inverted on the opposite side of the lens.

2. What is the formula for calculating the image position of a convergent lens?

The formula for calculating the image position of a convergent lens is: 1/f = 1/o + 1/i, where f is the focal length of the lens, o is the object distance from the lens, and i is the image distance from the lens.

3. How does the object distance affect the image position in a convergent lens?

The object distance determines how far the object is from the lens, and it directly affects the image position. The closer the object is to the lens, the further the image will be from the lens. Similarly, the further the object is from the lens, the closer the image will be to the lens.

4. Can convergent lenses produce both real and virtual images?

Yes, convergent lenses can produce both real and virtual images depending on the location of the object relative to the lens. A real image is formed when the object is placed beyond the focal point, while a virtual image is formed when the object is placed between the lens and the focal point.

5. How can I determine the size of the image formed by a convergent lens?

The size of the image formed by a convergent lens can be determined using the magnification formula: M = -i/o, where M is the magnification, i is the image distance, and o is the object distance. A positive magnification indicates an upright image, while a negative magnification indicates an inverted image.

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