Converging and diverging lenses

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Homework Statement

a) A converging lens with a focal length of 15cm has an object 5cm tall placed 40cm to the left of it. To the right, 30 cm away, is a converging lens of focal length 10cm. use the thin lens and magnification formulae to locate the finale image. Find its magnification.

b) Repeat part A for a -10cm diverging lens.

Homework Equations

1/f = 1/di + 1/do
m = -(di/do)
m = hi/ho

The Attempt at a Solution

i tried using a scaled diagram but I am still pretty lost...please help!

 

[alphysicist]

Once you find the image of the first lens (treating it like any other single lens problem), then the image of the first lens becomes the object of the second lens, and so you can find its image (which will be the final image). You might have to be careful of the sign conventions.

For the total magnification, you find the magnification of each lens just as in a single lens problem. Once you have the separate magnifications, what would be the total magnification?

 

[Moetasim]

I would suggest approaching this problem by first understanding the basics of converging and diverging lenses. A converging lens is thicker in the middle and causes parallel rays of light to converge at a point, while a diverging lens is thinner in the middle and causes parallel rays of light to spread out.

For part a) of the problem, we are given a converging lens with a focal length of 15cm and an object 5cm tall placed 40cm to the left of it. We are also given another converging lens with a focal length of 10cm, 30cm to the right of the first lens.

To solve this problem, we can use the thin lens formula: 1/f = 1/di + 1/do, where f is the focal length, di is the image distance, and do is the object distance. We can also use the magnification formula: m = -(di/do), where m is the magnification, di is the image distance, and do is the object distance.

First, we can calculate the image distance for the first lens using the thin lens formula. Plugging in the values, we get 1/15 = 1/di + 1/40. Solving for di, we get di = 24cm. This means that the image formed by the first lens is 24cm to the right of the lens.

Next, we can use the magnification formula to calculate the magnification of the first lens. Plugging in the values, we get m = -(24/40) = -0.6. This means that the image formed by the first lens is inverted and 0.6 times the size of the object.

Now, we need to find the image distance for the second lens. We can do this by using the thin lens formula again, but this time we will use the image distance from the first lens (24cm) as the object distance for the second lens. So, 1/10 = 1/di + 1/24. Solving for di, we get di = 15.43cm. This means that the final image is formed 15.43cm to the right of the second lens.

To find the overall magnification, we can use the fact that the magnification of a compound system of lenses is equal to the product of the individual magnifications. So, the overall magnification is (-0