Two convex lenses of focal lengths \(20 \mathrm{cm}\) and 10 \(\mathrm{cm}\) are placed \(30 \mathrm{cm}\) apart, with the lens with the longer focal length on the right. An object of height \(2.0 \mathrm{cm}\) is placed midway between them and observed through each lens from the left and from the right. Describe what you will see, such as where the image(s) will appear, whether they will be upright or inverted and their magnifications.

Let's identify the variables and given information in the problem: 1. Focal lengths of the two lenses: \[F1 = 20 \thinspace cm\] and \[ F2 = 10 \thinspace cm\] 2. Distance between the lenses: \[D = 30 \thinspace cm\] 3. Object distance : \[u = \frac{D}{2} = 15 \thinspace cm\] 4. Object height: \[h_o = 2.0 \thinspace cm\]

We will now calculate the image positions for each lens using the lens formula: \[{1 \over f} = {1 \over v} - {1 \over u}\] For Lens 1 (\[F1 = 20\thinspace cm\]): \[ {1\over 20} = {1 \over v_1} - {1\over 15}\] \[ {1 \over v_1} = {1\over 20} + {1\over 15}\] \[ {1 \over v_1} = {35 \over 300}\] \[ v_1 = {300\over 35} \thinspace cm \] For Lens 2 (\[F2 = 10\thinspace cm\]): \[ {1\over 10} = {1 \over v_2} - {1\over 15}\] \[ {1 \over v_2} = {1\over 10} + {1\over 15}\] \[ {1 \over v_2} = {5 \over 150}\] \[ v_2 = {30 \thinspace cm}\]

Next, we will calculate the magnification and determine the image orientation for each lens using the magnification formula: \[M = -{v \over u}\] For Lens 1 (\[v_1 = \frac{300}{35} \thinspace cm\]): \[M_1 = -{\frac{300}{35} \over 15}\] Since \(M_1\) is negative, the image will be inverted. For Lens 2 (\[v_2 = 30 \thinspace cm\]): \[M_2 = -{30 \over 15}\] Since \(M_2\) is negative, the image will also be inverted.

Finally, we will describe the image positions and properties based on our findings from Steps 2 and 3: When observing the object through the lens with focal length 20 cm (Lens 1) from the left, we will see an inverted image formed at a distance of \(\frac{300}{35}\) cm to the right of the lens. Similarly, when observing the object through the lens with focal length 10 cm (Lens 2) from the right, we will see an inverted image formed at a distance of 30 cm to the left of the lens. Regarding magnification, Lens 1 will produce a magnification of \(-{\frac{300}{35} \over 15}\), while Lens 2 will produce a magnification of \(-{30 \over 15}\). It is important to note that the magnifications are negative, meaning that the images formed by both lenses will be inverted.