A 1.20-cm -tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm. A

Step 1: Calculate the image distance and image height for the first lens with a focal length of

A 1.20-cm -tall object is 50.0 cm to the left of a...

A $1.20-\mathrm{cm}$ -tall object is $50.0 \mathrm{~cm}$ to the left of a converging lens of focal length $40.0 \mathrm{~cm}$. A second converging lens, this one having a focal length of $60.0 \mathrm{~cm},$ is located $300.0 \mathrm{~cm}$ to the right of the first lens along the same optic axis. (a) Find the location and height of the image (call it $I_{1}$ ) formed by the lens with a focal length of $40.0 \mathrm{~cm}$. (b) $I_{1}$ is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

Transcript

00:01 Hi there, so for this problem, we're told that a 1 .20 centimeters tall object, so we are given its height 1 .2 centimeters, and it is positioned at 50 centimeters to the left of a converging lens, so the object distance is given, and that is equal to 50 centimeters.

00:29 And with a focal length that is also given.

00:32 Now, since the length, are converging, this focal length is positive and that is equal to 40 centimeters.

00:41 And now there is also a second converging lens.

00:47 This is having a focal length that is we're going to call that the focal length f prime and that is equal to 60 centimeters.

00:56 And it is located at a distance of 300 centimeters to the right of the first length along the same opposite.

01:04 So the first question for this is to find the location of an height of the image formed by the lens with a focal length of 40 centimeters.

01:17 So that means this first image.

01:21 So what we need to determine is the image distance and the object's height and the image height, sorry.

01:29 So to obtain the image distance we use the thin length equation that states that the inverse of the focal length is equal to the inverse of the object distance plus the inverse of the image distance.

01:45 So if we solve for the image distance, that will be just simply the inverse of the focal distance minus the inverse of the object distance and all of that to the minus one.

01:57 So now we substitute the values 1 divided by the focal distance that is equal to 40 centimeters.

02:03 This minus the inverse of the image distance that is 50, and all of that to the minus 1.

02:11 So from this we obtain that the image distance is equal to a value of 200 centimeters.

02:30 So now recall that since this is on converging lens, this distance is going to be in the opposite side, of the object.

02:41 So with that said, let me just draw this.

02:44 Well, let's first calculate the image height.

02:48 So the image height, and to obtain that, we can use the definition of the magnification.

02:55 The magnification is that this is minus the image distance divided by the object distance, but this is also equal to the image height divided by the object's height.

03:05 So if we solve for the objects and for the image distance, we will obtain that there is the object's height minus the image distance divided by the object distance, and now we substitute the values.

03:18 So the object's height is 1 .2 centimeters, that times minus the image distance, which we obtain is 200 centimeters, that divided by the object's distance that is 50 centimeters.

03:40 So from this, we obtain a value of minus 4 .8 centimeters.

03:53 So that is the image height.

03:56 The minus sign indicates that the image is inverted.

04:01 So this is the image distance, the image distance, and the image height.

04:08 Now, with that said, we are now in the condition where we have the following.

04:14 So let's just draw in here the two lens.

04:19 This is the first length.

04:20 This is the second lens.

04:22 So something like this...