A converging lens with focal length 3.00 cm is placed 4.00 cm to the right of an object. A div

step 1 Let's start by writing down the information available to us. So we have that the focal length of

A converging lens with focal length 3.00 cm is placed 4.00...

A converging lens with focal length $3.00 \mathrm{~cm}$ is placed $4.00 \mathrm{~cm}$ to the right of an object. A diverging lens, with focal length $-5.00 \mathrm{~cm}$ is placed $17.0 \mathrm{~cm}$ to the right of the converging lens. (a) At what location(s), if any, can you place a screen in order to display an image? (b) Repeat part (a) for the case where the lenses are separated by $10.0 \mathrm{~cm}$.

A converging lens with focal length $3.00 \mathrm{~cm}$ is placed $4.00 \mathrm{~cm}$ to the right of an object. A diverging lens, with focal length $-5.00 \mathrm{~cm}$ is placed $17.0 \mathrm{~cm}$ to the right of the converging lens. (a) At what location(s), if any, can you place a screen in order to display an image?
(b) Repeat part (a) for the case where the lenses are separated by $10.0 \mathrm{~cm}$.

Key Concepts

Sign Conventions in Optics

Proper application of sign conventions is crucial when using the thin lens equation. These conventions determine the positive and negative values for object distances, image distances, and focal lengths, distinguishing between real and virtual images and ensuring accurate analysis of optical setups.

Lens Separation

The distance between lenses in a multiple-lens system plays a significant role in image formation. It affects how intermediate images are transferred from one lens to the next, influencing the final image's position, clarity, and magnification, which is critical for designing and understanding complex optical setups.

Real and Virtual Images

Real images are produced when light actually converges and can be projected onto a screen, whereas virtual images occur when light rays only appear to converge but cannot be directly captured. Recognizing the difference between these two types is essential for determining practical image formation and screen placement in optical systems.

Thin Lens Equation

The thin lens equation, given by (1/object distance) + (1/image distance) = 1/focal length, is fundamental in geometrical optics. It is used to calculate the position, size, and nature of an image produced by a thin lens, whether it be converging or diverging.

Sequential Imaging with Multiple Lenses

In systems involving more than one lens, the image from the first lens serves as the object for the next. This sequential imaging requires the step-by-step application of the thin lens equation, making it necessary to account for the distance between lenses to correctly trace the path and properties of light through the entire system.

Transcript

00:01 Let's start by writing down the information available to us.

00:04 So we have that the focal length of our first converging lens is three centimeters.

00:14 And the distance from our object to that first lens is four centimeters.

00:21 We also know that the focal length of our second lens is negative five centimeters because it's a diverging lens.

00:30 And we know that the separation between the two lenses, the distance between them, is 17 centimeters.

00:39 Now, we want to know how far behind that first lens are we going to get our image formed.

00:49 So we have our object here, we have our focal points, focal lengths, and we want to know how far away are we going to form this object here.

01:05 Q1, this length, how long is that? over here we have p1.

01:09 We can use the thin lens formula, which is 1 over f equals 1 over p plus 1 over q to solve for that.

01:19 If we rearrange it, we get that q1 equals 1 over f1 minus 1 over p1 to the negative 1.

01:29 So everything is going to be 1 over the stuff in that brackets.

01:32 And if you plug in the values you have, you get that q1 is 12 centimeters.

01:40 So this is 12 centimeters away.

01:43 Now how much further until we hit that second diverging lens? well, s is 17 centimeters, so s minus q1 minus 12, will give us 5 centimeters, which is also the distance from our new object or our real image produced by our first lens to our second lens.

02:04 So that's going to be p2.

02:08 This happens to be the focal length of our second lens.

02:13 But regardless, we now have a real image that's going to be going through a diverging lens.

02:22 So all of the libraries are going to diverge from here, and we're going to have only a virtual image.

02:29 So we can't put a screen down and view that image.

02:33 So the only real image in this problem right now is that first one...

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