In the figure, an object is placed in front of a converging...

In the figure, an object is placed in front of a converging lens at a distance equal to twice the focal length f1 of the lens. On the other side of the lens is a concave mirror of focal length f2 separated from the lens by a distance 2(f1 + f2). Light from the object passes rightward through the lens, reflects from the mirror, passes leftward through the lens, and forms a final image of the object. Take f1 = 5.4 cm and f2 = 3.9 cm. What are (a) the distance between the lens and the final image and (b) the overall lateral magnification M of the object? Is the image (c) real or virtual (if it is virtual, it requires someone looking through the lens toward the mirror), (d) to the left or right of the lens, and (e) inverted or noninverted relative to the object?

Transcript

00:01 Here we have a problem of optics.

00:04 We take some converging lens of focal length f1 and we take some concave mirror of focal length f2 and we put this image the distance 2 times f1 to the left of the lens.

00:20 And this lens and the mirror are separated by a distance 2 times f1 plus 2 times f2.

00:28 So we are going to get the final image together with its properties, namely its distance, its magnification and its realness, and also whether it is inverted or non -inverted.

00:45 So let's get started.

00:51 Due to the lens, we will have the first image.

00:57 First image due to lens.

01:01 So we will have the following sub -configuration, let's say.

01:11 We have this object here, the distance to f1 from the lens.

01:20 And we write down the thin lens equation.

01:24 One over the object distance plus one over the image distance equal to one over the focal length.

01:33 And since we have a converging lens, the convention.

01:38 Tells us that this focal length should be taken to be positive.

01:44 So let us write down what we know.

01:46 We have the object at a distance to f1.

01:49 We don't know the image distance.

01:52 And we have the right hand side.

01:55 And solving this equation for i, we get i equals 2f1.

02:03 Now, the sign conversion tells us that since we have a positive distance for a positive distance for the image it should be on the side opposite to the object namely it will be here on the other side and let us have a look at the magnification so let's call this m1 we have minus i over o minus so the image distance is 2 f1 and the object distance is again 2 f1 so we see that the magnification of this first image due to the lens will be minus one namely it will be inverted with respect to this object but they will have the same size so there will be no immediate magnification of this object as an image now we have this second image due to the middle image due to mirror and it will be as follows so the mirror is here and since this lens and the mirror are separated by distance to f1 plus 2f2 we will see that and let us put it the coin place it will inverted like this so now this first image will behave as an object for the mirror.

04:14 It is inverted to begin with it is at a distance to f2 from the mirror and the important point is that this image due to the lens is in front of the mirror so it will behave as a real image a real object behave as a real object for this mirror.

04:44 Now let us write down the image, the reflection equation again.

04:53 We have this one over the object distance plus the one of the image distance equal to one over the focal length of the mirror.

05:02 Again, the sign common, the usual sign convention tells us that since we have a concave mirror, we should have this f2 value.

05:13 Greater than zero now we have the object at a distance to f2 plus we have this image unknown image distance equals 1 hour f2 and so into the equation for i we get 2 f2 now for the mirrors we have this flipped sign convention namely since we have this positive image distance, image will be on the same side as the object.

06:10 So it will be a real image.

06:16 And we can also emphasize again that in the first case we have a real image again.

06:25 So it has just emphasized this for the sake of completeness.

06:30 Okay, now let us have a look at the magnification of this second image...

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