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. $\begin{array}{lllll}\mathbf{8 3} & +20 & \text { C } 9.0 & 8.0 & \text { C }, 5.0\end{array}$

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.
$\begin{array}{lllll}\mathbf{8 3} & +20 & \text { C } 9.0 & 8.0 & \text { C }, 5.0\end{array}$

Key Concepts

Inversion of Images

The concept of image inversion is related to the orientation of the image relative to the object. An inverted image is one that is upside down compared to the object, which can be determined by the sign of the magnification. This is important for diagnosing the behavior of images produced by lens systems and understanding how optical instruments manipulate visual information.

Real and Virtual Images

Real images are formed when light rays actually converge, meaning the image can be projected on a screen, whereas virtual images occur when the light rays do not actually meet but only appear to do so when traced backwards. This distinction is essential for understanding the output of a lens system and for proper application of the thin lens formula.

Sign Conventions

Sign conventions are rules for assigning positive or negative values to object distances, image distances, and focal lengths. They are critical in applying formulas consistently, determining the nature of images (such as real or virtual), and ensuring the correct interpretation of the orientation and position of the formed image.

Magnification and Lateral Magnification

Magnification in optical systems quantifies how much larger or smaller the image is compared to the object and includes information about the orientation of the image. Lateral magnification is calculated as the ratio of the image height to the object height, and it can be derived for each lens separately, with the overall magnification in a compound system being the product of the individual magnifications.

Compound Lens System

A compound lens system involves more than one lens, where the image from one lens becomes the object for the next. This concept is crucial in understanding how multiple optical elements work in sequence to form a final image, taking into account the distances between the lenses and the object, as well as between the lenses themselves.

Thin Lens Equation

The thin lens equation relates the object distance, image distance, and the focal length of a lens through the formula 1/f = 1/p + 1/i. This equation is fundamental to understanding how lenses form images and is used to calculate where an image will form given a particular object placement and lens type.

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 botset regions, as is shown in the figure at the right.

00:16 Now lens 1 is mounting within the bouttset region closer to the object, which is at an object distance b1.

00:27 And lems 2 is mounted within the farther bot set region at a distance d.

00:35 Now, the information for this problem is provided in this table.

00:40 Since we are working with problem 83, we are given the object distance 1, the type of lens 1, its focal distance, the distance d, the type of lens 2, and its focal distance.

00:54 Now for part a of this problem, we need to find the image distance for the image produced by the lens 2, that is the final image produced by this system that we called i2.

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

01:26 So we note that the image distance 1 is equal to the product between the object distance 1 at the focal distance 1 overt the difference between these two values.

01:39 So we note that the object distance 1 and that value is given in the table and that is a value of 20 centimeters.

01:50 And since we are working with converging lens, the focal distance 1 is equal to a positive value of 9 centimeters.

02:01 So we substitute those two values into this equation, and we will obtain that the image distance 1 is equal to 16 .4 centimeters.

02:19 Now, to obtain the image distance 2, we know that there is a product between the object distance 2 times the focal distance 2 over the difference between these two expressions.

02:31 So the focal distance 2 is given in the table above.

02:36 That value is positive since we are working again with converging length, and that has a value of 5 centimeters.

02:45 Now, the object distance 2 is equal to the distance d minus the object distance 1, sorry, the image distance 1.

02:55 So we substitute those two values.

02:58 The distance d is given in the table, 8 centimeters, minus the image distance 1, that is 16 .4 centimeters...

Please give Ace some feedback