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. $\begin{array}{lllll}\mathbf{8 7} & +20 & \mathrm{D}, 12 & 10 & \mathrm{D}, 8.0\end{array}$

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.
$\begin{array}{lllll}\mathbf{8 7} & +20 & \mathrm{D}, 12 & 10 & \mathrm{D}, 8.0\end{array}$

Key Concepts

Thin Lens Equation

This is the fundamental relation in geometric optics that connects the focal length of a lens with the object and image distances. It is commonly written as 1/f = 1/p + 1/i, where p is the object distance and i is the image distance. Correct application of this equation is essential in locating the images formed by individual lenses.

Lateral Magnification

Lateral magnification describes how much larger or smaller the image is compared to the object, and it also indicates whether the image is inverted. It is calculated as the negative ratio of the image distance to the object distance (M = -i/p) for thin lenses, with the sign indicating the orientation of the image.

Two-Lens Systems

In systems containing two lenses, the image formed by the first lens acts as the object for the second lens. Analyzing these systems involves applying the thin lens equation sequentially for each lens, taking into account the separation distance between them. This concept is key in understanding more complex optical instruments.

Converging vs Diverging Lenses

Converging (convex) lenses and diverging (concave) lenses behave differently in that converging lenses bring parallel rays to a focal point while diverging lenses spread them out. Identifying the type of each lens is crucial because it affects the sign of the focal length and the resulting image characteristics.

Real vs Virtual Images

Real images are formed when light rays actually converge, allowing the image to be projected on a screen, while virtual images occur when the rays diverge and the image only appears to form at a location. Recognizing whether an image is real or virtual is important when analyzing optical setups, especially in systems with multiple lenses.

Sign Conventions in Optics

The correct use of sign conventions is critical in applying the thin lens equation and determining magnification. This includes assigning proper signs to object distances, image distances, and focal lengths, which directly influence the interpretation of the final image's location, orientation, and nature in optical systems.

Transcript

0:00 Hi there.

00:01 So for this problem we have a stick figure that stands 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.

00:15 Now lens 1 is mounted within the bot set region that is closer to the object which is at an object distance p1.

00:25 Meanwhile, the lens 2 is mounting within the farther bot set region at a distance d.

00:33 Now, the information for this problem is provided in this table.

00:38 Since we are working with problem 87, we are given the object distance, the type of lens 1, its focal distance, the distance d, the type of lens 2, and its focal distance.

00:54 Now the first question for this problem is is to calculate the image distance for the image produced by the lens 2, that is the final image produced by this system that we call i2.

01:10 Now to calculate this quantity, we need to first calculate the image distance 1, because that serves as the object for the lens 2.

01:20 Now we note that that image distance is equal to the product between the object distance one and the focal distance one overt the difference between these two quantities.

01:33 Now, the focal distance one is given in the table, and it is a negative value because we are working with diverging lens, and that is minus 12 meters.

01:52 The object distance is also given, and that is a positive value of 20 centimeters.

02:02 So after substituting these values into the expression, we will obtain that the image distance 1 is equal to minus 7 .5 centimeters.

02:21 Now we proceed to calculate the image distance 2 that we know is the product between the object distance 2, the focal distance 2 over the difference between these 2.

02:40 Now, in the table we are given the focal distance 2 that is a positive, sorry, it is a negative value because we are working with diverging lens again.

02:56 So that is a negative value of 8 centimeters...

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