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. $\begin{array}{lllll}86 & +12 & \text { C, } 8.0 & 30 & \text { D, } 8.0\end{array}$

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.
$\begin{array}{lllll}86 & +12 & \text { C, } 8.0 & 30 & \text { D, } 8.0\end{array}$

Key Concepts

Lateral Magnification

Lateral magnification quantifies the ratio of the image size to the object size, often expressed as m = -i/p for a single lens. In multi-lens systems, the overall magnification is the product of the individual magnifications produced by each lens, making it an important concept in understanding how size and orientation are altered in compound optical systems.

Sign Conventions

Sign conventions in optics dictate the positive and negative values for distances and focal lengths. For instance, real images are typically given positive image distances while virtual images have negative ones. These conventions are essential for correctly applying the lens equation and determining the nature of the image (real or virtual, inverted or upright).

Compound Lens Systems

In a compound or multi-lens system, the image formed by the first lens serves as the object for the second lens. Analyzing such systems requires sequential application of the lens equation and magnification formula, highlighting the need to consider intermediate images and their properties to determine the final image's characteristics.

Lens Equation

This equation, commonly expressed as 1/p + 1/i = 1/f, is fundamental for analyzing image formation in lenses. It relates the object distance (p), the image distance (i), and the focal length (f) of the lens, allowing one to determine where an image is formed relative to the lens.

Thin Lens Approximation

The thin lens approximation assumes that the lens has negligible thickness compared to the object and image distances. This simplification permits the use of the lens equation without accounting for the lens thickness, which is crucial when analyzing systems like the two-lens system in the problem.

Transcript

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

00:18 Now lens one is mounted within the bot set region closer to the object which is at an object distance b1 and the lens 2 is mountain within the farther botset region at a distance d.

00:33 Now the information for this problem is provided in this table and since we are working with problem 80 sets we are given the image the the object distance 1, the type of lens 1, its focal distance the distance d, the type of lens 2 and its focal length.

01:02 Now for part a of this problem, we are asked about the image distance for the image that is produced by the lens 2.

01:11 That means the final image produced by this whole system.

01:18 In that image, we call it image 2.

01:22 Now to obtain this value, we need first to obtain the image distance 1 because that serves as the object for the lens 2.

01:33 Now the image distance 1 is equal to the product between the object distance 1 and the focal distance 1 overt the difference between these 2.

01:48 Now the object distance 1 is given in this table and that is a positive value of 12 centimeters and the focal distance 1 is a positive value since the lens 1 is a converging length, so that is 8 centimeters.

02:09 So what we need to do is to substitute those values into this expression.

02:14 And we will obtain that the image distance 1 is a positive value of 24 centimeters.

02:23 Now we proceed to obtain the image distance 2, which is the object distance 2 times the focal distance 2.

02:33 And this over the object distance 2 minus the focal distance 2.

02:41 Now, the focal distance 2, that value is given in the table, and since the type of lens is diverging lens, we will have that this is a negative value of 8 cm.

02:57 And the object distance 2, we can obtain that from the distance d minus the image distance 1.

03:06 So substituting those values in here, the value of the is given in the table, and that is a value of 30 centimeters, minus the image distance 1, that is 24 centimeters...

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