95 through 100. 95, 96, 99. Three-lens systems. In Fig....

95 through 100. 95, 96, 99. Three-lens systems. In Fig. $34-49$ , stick figure $O$ (the object) stands on the common central axis of three thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closest to $O$ , which is at object distance $p_{1} .$ Lens 2 is mounted within the middle boxed region, at distance $d_{12}$ from lens $1 .$ Lens 3 is mounted in the farthest boxed region, at distance $d_{23}$ from lens $2 .$ Each problem in Table $34-10$ refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by C for converging and D for diverging; the number after $\mathrm{C}$ or $\mathrm{D}$ is the distance between a lens and either of the focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance $i_{3}$ for the (final) image produced by lens 3 (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 $(\mathrm{V}),$ (d) inverted $(1)$ from object $O$ or noninverted $(\mathrm{NI}),$ and $(\mathrm{e})$ on the same side of lens 3 as object $O$ or on the opposite side.

95 through 100. 95, 96, 99. Three-lens systems. In Fig. $34-49$ , stick figure $O$ (the object) stands on the common central axis of three thin, symmetric lenses, which are mounted in the boxed
regions. Lens 1 is mounted within the boxed region closest to $O$ ,
which is at object distance $p_{1} .$ Lens 2 is mounted within the middle boxed region, at distance $d_{12}$ from lens $1 .$ Lens 3 is mounted in the farthest boxed region, at distance $d_{23}$ from lens $2 .$ Each problem in Table $34-10$ refers to a different combination of lenses and
different values for distances, which are given in centimeters. The type of lens is indicated by C for converging and D for diverging; the number after $\mathrm{C}$ or $\mathrm{D}$ is the distance between a lens and either of the focal points (the proper sign of the focal distance is not
indicated).
Find (a) the image distance $i_{3}$ for the (final) image produced by lens 3 (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 $(\mathrm{V}),$ (d) inverted $(1)$ from object $O$ or noninverted $(\mathrm{NI}),$ and $(\mathrm{e})$ on the same side of
lens 3 as object $O$ or on the opposite side.

Key Concepts

Image Characteristics (Real vs Virtual, Inverted vs Noninverted)

The characteristics of the final image—such as whether it is real or virtual, and whether it is inverted or upright—depend on the nature of the lens and the computed image distance and magnification. A real image, formed where light converges, can be captured on a screen, while a virtual image, appearing behind the lens, cannot. The sign of the magnification determines the image’s orientation: a negative sign indicates inversion relative to the object, whereas a positive sign signifies the same orientation.

Magnification

Magnification in optics quantifies how much larger or smaller an image is compared to the object, and for a single lens, the lateral magnification is given by m = -i/p. In multi-lens systems, the overall magnification is the product of the magnifications of the individual lenses. This concept helps determine not only the size of the image but also its orientation relative to the object, indicating whether the image is inverted or not.

Ray Diagrams and Sign Conventions

Ray diagrams are visual representations used to trace the path of light through lenses, while sign conventions assign positive or negative values to object distances, image distances, and focal lengths. These conventions are essential for accurately applying the thin lens equation and for determining characteristics such as whether an image is real or virtual, as well as its orientation. Mastery of these sign rules ensures consistency across different optical systems.

Thin Lens Equation

The thin lens equation, 1/f = 1/p + 1/i, relates the focal length of a lens to the object distance (p) and the image distance (i). It is fundamental in geometric optics and is used to determine where an image forms with respect to a thin lens. Understanding this equation is critical because it forms the basis for analyzing single-lens systems before extending the concepts to more complex, multi-lens arrangements.

Multi-Lens Systems

In multi-lens systems, such as three-lens configurations, each lens forms an image that often serves as the object for the next lens. Analyzing such systems requires sequential application of the thin lens equation and careful consideration of the distances between lenses. This approach is crucial in fields like optical engineering where complex devices, such as telescopes and microscopes, rely on multiple lens elements to form a final image.

Transcript

00:01 For part a of our question, we're asked to find the distance of the image produced by the second, or the third lens by the third lens.

00:07 So to do that, we're going to find the distance, or to do that, we need to find the object distance for the third lens, since we know it's focal length.

00:15 The object distance for the third lens will be the image produced by the second lens, and the object distance for the second lens will be the image produced by the first lens.

00:22 So we'll use the equation 1 over i sub 1 plus 1 over p sub 1 is equal to 1 over the focal length of lens 1 to solve for the image distance of the first lens.

00:30 Lens i sub 1.

00:32 And we find that this is equal to 1 over f sub 1 minus 1 over p sub 1, and everything there needs to be raised to the negative 1 to get i sub 1 from the denominator back to the numerator.

00:46 Plugging in these values, we find that this is equal to 9 centimeters.

00:51 Now we can use that to find the distance to the object of the second lens, p sub 2.

00:56 It's going to be the distance between the lens of 1 and 2, d sub 1 and 2, minus the image of the first lens.

01:03 Plugging in these values, we find that this is equal to six centimeters.

01:08 Now that we know this, we can find i .2 in the same way we found i sub 1, but we're going to replace everything in the brackets with the values for the second lens.

01:17 So this is 1 over f sub 2 minus 1 over p sub 2.

01:24 Raising all this to the negative 1, we find that i sub 2 is equal to 6 centimeters.

01:33 Now we can use the value for i sub 2 to find p.

01:37 Sub 3 in a similar manner to p.

01:38 Sub 2.

01:39 Except this time we're using the distance between the second and the third lens minus i sub 2.

01:45 We find that this is equal to 5 centimeters, and now we can find i sub 3, where i sub 3 is equal to 1 over f sub 3 minus 1 over p sub 3...