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. $\mathbf{8 2} \quad+8.0$ $\begin{array}{lll}\text { D. } 6.0 & 12 & \text { C, } 6.0\end{array}$

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.
$\mathbf{8 2} \quad+8.0$
$\begin{array}{lll}\text { D. } 6.0 & 12 & \text { C, } 6.0\end{array}$

Key Concepts

Thin Lens Equation

The thin lens equation (1/f = 1/p + 1/i) is fundamental in relating the focal length of a thin lens with the object distance (p) and image distance (i). It serves as the basis for analyzing how lenses form images and is applicable in both converging and diverging lenses, provided proper sign conventions are used.

Sign Conventions

Sign conventions in optics provide a systematic way to assign positive or negative values to distances and focal lengths. They help determine the nature (real or virtual) and orientation of images formed by lenses. These conventions are essential when applying the thin lens equation to ensure consistency and accuracy in image analysis.

Image Magnification

The magnification produced by a lens is given by the ratio -i/p, which shows how the size of the image compares to the size of the object and can indicate whether the image is inverted or not. In multi-lens systems, overall magnification is found by multiplying the magnification produced by each lens, reflecting the cumulative effect on the object's image.

Compound Lens Systems

When dealing with a system that contains more than one lens, each lens is analyzed sequentially. The first lens forms an image that acts as the object for the second lens. This step-by-step approach is critical for accurately determining the final image position and characteristics in compound optical systems.

Real vs. Virtual Images

The determination of whether an image is real or virtual depends on the sign of the image distance as dictated by the sign conventions. Real images are formed when light rays converge, while virtual images appear to diverge from a point. Understanding this distinction is key when tracing rays through optical systems.

Image Orientation

Image orientation is inferred from the sign of the magnification. A negative magnification indicates that the image is inverted relative to the object, while a positive magnification indicates that it remains upright. In compound systems, the overall orientation is determined by the product of the individual magnifications from each lens.

Transcript

00:01 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 butt -set regions as is shown in the figure at the right.

00:17 Now, lens 1 is mounting within the bot set region that is closer to the object, which is at the object distance p1.

00:28 The lens 2 is mounting within the farther bot set region that is at a distance d.

00:36 Now the information for this problem is provided in this table.

00:41 Since we are working with problem 82, we are given the object distance 1, the type of lens 1, its focal length, the distance d, the type of lens 2, and its focal lens.

00:56 And so for part a of this problem, what we need to calculate or determine is the image distance for the image produced by the lens 2 that we called i2.

01:13 Now, to obtain this quantity or this value, the first thing that we need to calculate is the image distance 1, because that image serves as the object for the lens 2.

01:32 So to obtain the image distance 1, we know that that is a product between the object distance 1, the focal distance 1 overt the difference between these 2.

01:46 Now, we are given that the focal distance 1, it is 6 centimeters.

01:53 However, since we are working with a diverging lens, that means that the focal distance distance must be negative, so we will have minus 6 centimeters.

02:05 And the object distance 1 is equal to 8 centimeters.

02:11 Now we substitute these values into the equation, and then we will obtain that the image distance 1 is minus 3 .4 centimeters.

02:27 Now, we proceed to calculate this.

02:30 The image distance 2, that is the product between the object distance 2 and the focal distance 2 over the difference between these two quantities or values.

02:43 Now the value for the focal distance 2 is given in the table and that is a positive value since the lens 2 is a converging lens.

02:56 And so we are given the value that that is 6 centimeters.

03:03 And the object distance 2 is equal to the distance d minus the value that we obtain for the image distance 1.

03:12 So we just substitute those values in here.

03:14 Now the image, the distance d is 12, 12 centimeters minus 3 .4 centimeters...

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