Two-lens systems. In Fig. 34-45, stick figure O (the object)...

Two-lens systems. In Fig. $34-45$, stick figure $O$ (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to $O$, which is at object distance $p_{1}$. Lens 2 is mounted within the farther boxed region, at distance $d .$ Each problem in Table $34-9$ refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by $\mathrm{C}$ for converging and D for diverging; the number after C or $\mathrm{D}$ is the distance between a lens and either of its focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance $i_{2}$ for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnification $M$ for the system, including signs. Also, determine whether the final image is (c) real (R) or virtual (V), (d) inverted (I) from object $O$ or noninverted (NI), and (e) on the same side of lens 2 as object $O$ or on the opposite side. $\begin{array}{lll}84 & +15 & \text { C. } 12\end{array}$ 67 C. 10

Two-lens systems. In Fig. $34-45$, stick figure $O$ (the object) stands on the common central axis of two thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closer to $O$, which is at object distance $p_{1}$. Lens 2 is mounted within the farther boxed region, at distance $d .$ Each problem in Table $34-9$ refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by $\mathrm{C}$ for converging and D for diverging; the number after C or $\mathrm{D}$ is the distance between a lens and either of its focal points (the proper sign of the focal distance is not indicated).
Find (a) the image distance $i_{2}$ for the image produced by lens 2 (the final image produced by the system) and (b) the overall lateral magnification $M$ for the system, including signs. Also, determine whether the final image is (c) real (R) or virtual (V),
(d) inverted
(I) from object $O$ or noninverted (NI), and
(e) on the same side of lens 2 as object $O$ or on the opposite side.
$\begin{array}{lll}84 & +15 & \text { C. } 12\end{array}$
67
C. 10

Key Concepts

Thin Lens Equation

The thin lens equation, given by 1/p + 1/i = 1/f, relates the object distance (p), image distance (i), and focal length (f) for a thin lens. This fundamental equation is used to determine where an image will be formed by a lens, assuming that the lens’s thickness is negligible relative to the distances involved.

Sign Conventions

In optics, sign conventions assign positive or negative values to quantities such as object distances, image distances, and focal lengths. These conventions are essential for correctly identifying whether an image is real or virtual, its orientation, and its position relative to the lens. They ensure consistent and accurate application of the thin lens formula in different scenarios.

Compound (Two-Lens) Systems

A compound lens system involves two or more lenses arranged such that the image formed by the first lens serves as the object for the second lens. Analyzing such systems requires applying the thin lens equation to each lens sequentially and accounting for the distance between the lenses. This step-by-step approach allows one to determine the final image position and properties from the combination of lenses.

Magnification in Optical Systems

Magnification describes the ratio of the image size to the object size and, for a single thin lens, is given by m = -i/p. In multi-lens systems, the overall lateral magnification is found by multiplying the magnifications of the individual lenses. This parameter not only indicates the size change but also tells us about the inversion of the image, as a negative magnification implies an inverted image.

Image Characteristics (Real/Virtual and Inversion)

Images formed by lenses can be categorized as real or virtual. Real images are created when light rays converge and can be projected onto a screen, while virtual images form when rays appear to diverge from a point; such images cannot be captured on a screen. The determination of image orientation—whether an image is inverted or noninverted—is linked to the sign of the magnification and is a crucial aspect in the analysis of optical systems.

Transcript

00:01 Hi there, so for this problem we have a figure that stands on the common central axis of two thin symmetric lenses which are mounted in the bat set regions as is shown in the figure at the right.

00:15 Now lens 1 is mounted within the bat set region that is closer to the object which is in an object distance b1.

00:25 Now the lems 2 is mounted within the farther but set region at a distance d.

00:33 Now, the information for this problem is given in this table, and since we are working with problem 84, we are given the object distance 1, the type of lens 1, its focal distance, the distance d, the type of lens 2, and the focal distance 2, and the focal distance 2.

01:03 So the first question for this problem is to obtain the image distance for the image that is produced by the lems 2.

01:16 That is the final image produced by this system.

01:20 So we are called that the image distance 2.

01:25 Now to obtain this value, we need first to calculate the image distance 1 because that serves as the the object for the lens 2.

01:37 Now, the image distance 1 is defined as the product between the object distance 1, the focal distance 1, and the difference between these two quantities.

01:51 So the object distance 1 is given in the table, that is a value of 15 centimeters.

02:00 And the focal distance 1 is equal to 12 centimeters.

02:05 And this is a positive balance.

02:07 Since the type of lengths that we are working with are converging lengths.

02:13 Now we substitute these two values into this expression and we obtain that the image distance 1 is equal to the image is equal to a value of 60, a positive value of 60, sorry 60 centimeters.

02:46 Now for to obtain the image distance 2 we do the product between the object distance 2 the focal distance 2 over the difference between these 2 values now the focal distance 2 is given that value is given in the table and since again the lengths to our converging lens that means that the focal distance must be positive and that is a positive value of 10 centimeters and the object distance 2 is equal to the distance d minus the image distance 1, which is the distance d is equal to 67 centimeters minus the 60 centimeters for the image distance 1.

03:43 So from here we obtain that the object distance 2 is equal to 7 centimeters...

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