A converging lens (f=12.0 cm) is located 30.0 cm to the left of a diverging lens (f=-6.00 cm) .

step 1 So we have two lenses. Let's say this is L1 and let's say this is L2. The two lenses are 30 cent

A converging lens (f=12.0 cm) is located 30.0 cm to the left...

A converging lens $(f=12.0 \mathrm{cm})$ is located $30.0 \mathrm{cm}$ to the left of a diverging lens $(f=-6.00 \mathrm{cm}) .$ A postage stamp is placed $36.0 \mathrm{cm}$ to the left of the converging lens. (a) Locate the final image of the stamp relative to the diverging lens. (b) Find the overall magnification. (c) Is the final image real or virtual? With respect to the original object, is the final image (d) upright or inverted, and is it (e) larger or smaller?

Key Concepts

Image Orientation and Size

The orientation of an image (upright or inverted) is determined by the sign of the magnification; a negative magnification indicates an inverted image relative to the object, while a positive magnification indicates an upright image. Moreover, the magnitude of the overall magnification reveals whether the final image is larger or smaller than the object. Understanding these relationships is important for predicting how an optical system will alter image characteristics.

Real vs Virtual Image Formation

Real images are formed when light rays converge, and they can be projected onto a screen, whereas virtual images are formed when the light rays diverge and only appear to come from a specific location. The sign of the image distance resulting from the thin lens equation indicates whether an image is real (positive) or virtual (negative), making this concept crucial in determining the nature of the image produced.

Magnification

Magnification in lens systems is given by the ratio of the image distance to the object distance (with a negative sign for image inversion), usually expressed as m = -s'/s for a single lens. In multi-lens systems, the overall magnification is the product of the individual magnifications from each lens. This concept is key to understanding how the size and orientation of an image are transformed relative to the original object.

Sequential Imaging in Multi-Lens Systems

In systems with multiple lenses, the image produced by the first lens acts as the object for the subsequent lens. The process involves calculating the intermediate image position using the thin lens equation for the first lens, then using its image distance relative to the second lens (adjusted for the physical separation between lenses) to calculate the final image. This sequential approach is essential for tracing light paths and determining overall image properties in complex optical setups.

Thin Lens Equation

The thin lens equation, expressed as 1/s + 1/s' = 1/f, is fundamental in optics for relating the object distance (s), the image distance (s'), and the focal length (f) of a lens. This equation is used to determine where an image will form for a given object and lens configuration and is a critical tool for analyzing systems with one or more lenses.

Transcript

00:01 So we have two lenses.

00:03 Let's say this is l1 and let's say this is l2.

00:07 The two lenses are 30 centimeters apart and the focal length of l1 is equal to 12 centimeters positive because it's a converging lens and the focal length for the second lens is negative 6 .0 centimeters because it's a diverging lens.

00:29 And an object is placed 36 centimeters from l1.

00:35 So this is the original do, 36 .0 centimeters.

00:40 So part a says find the location of the final image.

00:45 So let's first talk about the lens l1.

00:48 So we have 1 over do plus 1 over d .i is equal to 1 over f, 1 over 36, plus 1 over d .i is equal to 1 over d .i, 1 over 12 and that gives me 1 over di is equal to 1 over 12 minus 1 over 36 and solving that for the i gives di is equal to 18 .0 centimeters so that means this image is to the right of l1 so this is where this image would be formed this distance here would be 18 centimeters and that would then mean that the distance of the object for lens l2 would then be 12 centimeters, which is the difference between 30 and 12.

01:42 So now we can find where the location of the image be for l2.

01:47 So for l2, we start with 1 over do plus 1 over d .i is equal to 1 over f and 1 over d .o.

01:57 Do is 12.

01:58 We just figured that out on the last screen, or plus 1 over d .i.

02:03 Is equal to negative 1 over 6.

02:06 So 1 over d .i is equal to negative 1 over 6 minus 1 over 12.

02:13 And that gives the i is equal to negative 4 centimeters.

02:19 So this image will be 4 centimeters to the left of l2...