An object is located 10.0 cm in front of a converging lens with focal length 12.0 cm. To the r

An object is located 10.0 cm in front of a converging lens...

Key Concepts

Thin Lens Equation

The thin lens equation, given by 1/f = 1/d_o + 1/d_i, is used to relate the focal length of a lens (f) to the distances of the object (d_o) and the image (d_i) from the lens. It is a fundamental tool in geometrical optics for determining the position at which an image is formed by a lens, assuming the lens is thin relative to the object and image distances.

Ray Tracing

Ray tracing is a method used to graphically determine the path taken by light as it passes through an optical system such as lenses. By drawing representative rays—like those parallel to the principal axis that pass through the focal point, or ones going through the center of the lens emerging undeviated—one can locate the position, orientation, and nature (real or virtual) of the resulting image.

Multiple Lens Systems

In multiple lens systems, the image formed by one lens acts as the object for the subsequent lens. Analyzing such systems typically involves sequentially applying the thin lens equation and ray tracing techniques to each lens. This approach is crucial for understanding complex optical devices, as it helps determine the final image characteristics and their locations based on the individual contributions of each lens.

Transcript

00:03 All righty, we've got this question right now and it has to do with a situation where you have two lenses.

00:09 They're both converging lenses.

00:11 So i'm going to make a line across here that we can set everything up on.

00:15 Now what we're going to do here is we're going to start off with the first lens.

00:21 And the first lens is situated, say right here, and it's got a focal length of 12.

00:30 And over here, so i'm going to put my two focuses in.

00:34 Here's one right here.

00:35 Let's call this 12 right here.

00:37 And they'll put another one over here at 12.

00:40 These are on either side, f1 and f2.

00:44 All right.

00:45 So now we have an object.

00:47 We're going to place that at 10.

00:49 So it's inside the focal length.

00:51 And it's going to be right about here.

00:53 Now, i'm also going to redraw this picture of this lens here so that it has a symbol.

00:58 So we can just see what happens if the lens is bigger than what i've, drawn it to be because we may need that for this problem.

01:08 All right, so let's draw some ray tracing to find out where will the image of this be.

01:13 We have three rays that we can use.

01:17 We can use the parallel ray.

01:19 So if we have a parallel ray, it's going to come back, it's going to travel through the focus.

01:27 Right? here it is over here.

01:30 And this can go in either direction, so we could have it coming over here also.

01:33 So there's the parallel ray.

01:37 Now we also have a, let's try this one in blue.

01:42 We have a center ray, which would go right through the center of the lens.

01:48 So if it went to the center of the lens, then it would continue like this.

01:53 And it turns out that these are diverging.

01:56 So this could also be seen as coming from this direction over here, and it will converge on the side over here.

02:03 The third one i want to put in the picture is going to be, i'll put it in red, is going to be my focal ray.

02:12 Now my focal ray, if i aim for the focus over here, there's no place that these three rays are going to converge.

02:20 So i'm going to use this focus over here and say my focal ray is going to be the one that goes through that focus.

02:28 And the focal ray comes back parallel, and so it comes back over here.

02:34 And this over here is where i get my image.

02:41 Here's my image over here.

02:42 Now, we're going to use this image as the object for the second ray, for the second lens.

02:48 But for now, let's just go ahead and using the formula for the lens equation to find out whether it's accurate what we've done here.

02:57 So we'll say 1 over f, equals one over object plus one over image.

03:04 And this is the image distance right here.

03:07 So the focus is 12.

03:10 1 over 12 equals 1 over 10 plus 1 over the image.

03:16 And we're going to use this to say, let's turn these into 60ths.

03:20 So that's going to be 5 over 60 equals 6 over 60.

03:27 And then this one is going to be plus.

03:29 Well, in order to go get to 5 over 60, we'll have to subtract minus 1 over 60.

03:35 And that's one over the image.

03:37 So the image distance is minus 60.

03:41 And this is very good because we have a minus 60, which means on the left side of the lens, if we started with the object on the left.

03:51 And that means this entire distance right here from the image to the lens is 60 centimeters.

03:59 Now the 60 centimeters here will also allow us to figure out what the magnification is.

04:05 Magnification is minus d .i.

04:08 Well, this is right this, minus image over object.

04:12 And so we have minus and minus 60.

04:15 So magnification is minus minus 60 is plus 60 over the object, which was 10 centimeters distance.

04:23 And so we get a magnification of six...

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