Question

Let an object be placed 2.0 cm in front of a converging lens of focal length 20.0 cm. A second converging lens of focal length 30.0 cm is placed 20.0 cm to the right of the first lens. Find the image location and its characteristics.

Answer 1

In a lens combination like this, first we have to find the image formed by the first lens, and then we'll treat this image as the object for the second lens.

Since the object is located in front of the focal point, a virtual image must be formed. This is because the light rays coming from the object diverge after passing through the lens, but if we trace backwards we can find the intersection point (the rays don't actually go through that point, that's why the image is called virtual). For every observer, the refracted rays would seem to be diverging from this point, therefore this is the image point. By using the lens equation we can determine the image exact position:

In a lens combination like this, first we have to find the image formed by the first lens, and then we'll treat this image as the object for the second lens. Since the object is located in front of the focal point, a virtual image must be formed. This is because the light rays coming from the object diverge after passing through the lens, but if we trace backwards we can find the intersection point (the rays don't actually go through that point, that's why the image is called virtual). For every observer, the refracted rays would seem to be diverging from this point, therefore this is the image point. By using the lens equation we can determine the image exact position: 1/f = 1/d_0 + 1/d_i rightarrow 1/20 = 1/2 + 1/d_i rightarrow 2 d_i = 20 d_i + 40 rightarrow d_i = -40/18 approx - 2, 2 m This is aligned with the

This is aligned with the sign conventions, since negative d_i means the image is located at the object's side of the lens, or virtual. The magnification can be calculated using

$M=-\frac{d_{i}}{d_{o}}=-\frac{-2,2cm}{2cm}=1,1$

And since it's positive and greater that 1, this means that image 1 is also upright and enlarged.

Now this image becomes the object for lens 2, and it's located 22,2 cm to the left of it. It's also going to be in front of the focal point, since focal length is 30 cm for lens 2. So we use the same equations again:

$\frac{1}{f}=\frac{1}{d_{o}}+\frac{1}{d_{i}}\rightarrow \frac{1}{30}=\frac{1}{22,2}+\frac{1}{d_{i}}\rightarrow 22,2d_{i}=30d_{i}+666\rightarrow d_{i}\approx -85,4cm$

$M=-\frac{d_{i}}{d_{o}}=-\frac{-85,4cm}{22,2cm}=3,85$

The resultant image is located 85,4 cm to the left of the second lens ( or 65,4 cm to the left of the first lens), therefore it's a virtual image, it's upright and magnified 3,85 times with respect to the first image, wich means it's upright with respect to the object as well and total magnification is

$M_{total}=M_{1}\times M_{2}=1,1\times 3,85=4,23$

this is, the resultant image is 4,23 times bigger that the object