A 2.0 -cm-tall object is 20 cm to the left of a lens with a...

A $2.0$ -cm-tall object is $20 \mathrm{cm}$ to the left of a lens with a focal length of $10 \mathrm{cm} .$ A second lens with a focal length of $-5 \mathrm{cm}$ is $30 \mathrm{cm}$ to the right of the first lens. a. Use ray tracing to find the position and height of the image. Do this accurately with a ruler or paper with a grid. Estimate the image distance and image height by making measurements on your diagram. b. Calculate the image position and height. Compare with your ray-tracing answers in part a.

A $2.0$ -cm-tall object is $20 \mathrm{cm}$ to the left of a lens with a focal length of $10 \mathrm{cm} .$ A second lens with a focal length of $-5 \mathrm{cm}$ is $30 \mathrm{cm}$ to the right of the first lens.
a. Use ray tracing to find the position and height of the image. Do this accurately with a ruler or paper with a grid. Estimate the image distance and image height by making measurements on your diagram.
b. Calculate the image position and height. Compare with your ray-tracing answers in part a.

Key Concepts

Magnification

Magnification is the measure of the relative size of the image compared to the object, defined as m = -di/do. The negative sign indicates image inversion, and its absolute value reveals whether the image is magnified or diminished relative to the object.

Sequential Lens Systems

In optical systems with multiple lenses, the image produced by one lens becomes the object for the next. Analyzing such systems requires sequential application of the thin lens equation and careful tracking of object and image distances between the components, ensuring that sign conventions and relative distances are correctly maintained.

Ray Tracing in Geometric Optics

Ray tracing is a graphical method used in optics to determine the formation and characteristics of images by drawing the propagation of selected rays through optical systems. It involves the use of principal rays—such as the ray parallel to the optical axis, the ray passing through the center of the lens, and the ray through the focal point—to accurately locate images and assess their size and orientation.

Thin Lens Equation and Sign Conventions

The thin lens equation, 1/f = 1/do + 1/di, connects the focal length (f) of a lens with the object distance (do) and the image distance (di), enabling the calculation of where an image will form. Applying proper sign conventions is critical for correctly identifying whether an image is real or virtual, and whether it is inverted or upright.

Transcript

00:01 Note that the first lens is a positive focal length, so it's a converging lens, and the second lens has a negative focal length, so it is a diverging lens.

00:10 Okay, let's do our rays for our first lens.

00:12 So here's our object, 27 meters to the left of our first lens.

00:16 First way we want to do is from the top of our object through the center of our first lens.

00:26 The second way we want to do is from the top of our object horizontal to our lens and then from there we can go to our focal point like so and you should find that we're going to get a ray or excuse me a new image a real image right there looks about 20 centimeters to the right okay we need to do the same thing for our second lens which is a diverging lens the first ray we want to do is from the center of our diverging lens through the top of our object.

01:16 Like so.

01:19 Next, we want to draw a ray to horizontally to our lens.

01:27 And then from there, we want to draw another ray through the focus on the same side like that.

01:40 We'll actually get our virtual, i wonder if i can zoom in to make this clear, virtual image right there.

01:53 Kind of small, but that's where it will be.

01:57 So what about this distance? this distance looks about at negative 3 centimeters from our second lens, and then our height of looks probably at around negative 1.

02:13 One or so.

02:16 But anyway, we can find out in part b just to be sure.

02:21 Okay, so we need to start with the equation 1 over focal length equals 1 over the position of the object plus one over the position of the image.

02:31 Let's start with the first lens...

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