(II) A diverging lens with f=-31.5 cm is placed 14.0 cm behind a converging lens with f=20.0 cm

step 1 In this problem, we have two lenses. One is a diverging lens with F is negative 31 .5. So F1 is

(II) A diverging lens with f=-31.5 cm is placed 14.0 cm...

Key Concepts

Composite Optical Systems

In a composite optical system, multiple lenses are arranged in sequence such that the image produced by one lens serves as the object for the next. This concept is important when analyzing complex setups where the overall behavior of the system is determined by the combined effects of individual lenses.

Diverging Lenses

Diverging lenses, characterized by negative focal lengths, spread out parallel rays of light as if they originated from a focal point on the same side of the lens as the incoming light. These lenses produce virtual, upright, and reduced images, and are crucial in systems that require image enlargement or correction for optical aberrations.

Sign Conventions in Optics

Correct application of sign conventions in optics is essential for accurately determining image locations and characteristics. These conventions assign positive or negative values to object distances, image distances, and focal lengths depending on the chosen coordinate system, ensuring consistency when applying the lens equation and other formulas.

Lens Equation

The lens equation (1/f = 1/d? + 1/d?) is fundamental in optics for determining the relationship between object distance, image distance, and focal length. It allows one to predict where an image will form given the position of the object relative to the lens, while taking into account the sign conventions for each parameter.

Converging Lenses

Converging lenses, which have positive focal lengths, are designed to focus parallel rays of light to a single point known as the focal point. They are often used in situations where one needs to create real, inverted images and are key components in many optical instruments.

Transcript

00:01 In this problem, we have two lenses.

00:03 One is a diverging lens with f is negative 31 .5.

00:08 So f1 is negative 31 .5 centimeters.

00:18 And it's 14 centimeters behind a converging lens.

00:26 F2 is equal to 20 centimeters.

00:43 And this distance is 14.

00:47 Centimeters and we have an object at infinity so do equals infinity and we want to know where it will be focused so let's find use the thin lens equation to find where the image is from the first lens so d -i is equal to the inverse of one over f minus one over do you've gotten this far in the chapter you probably know how to manipulate the thin lens equation to get it to look like this.

01:31 And we can plug in basically, oh, i guess we didn't really need to calculate.

01:41 If do is infinity, that goes away.

01:43 And then we just invert f.

01:45 We invert the inverse of f, and that just gives us f.

01:49 So f is the focal point is going to be where the image is focused.

01:56 And so then that means the image will be at negative 31 and a half.

02:02 So then that means it'll be behind this lens.

02:06 So let's call this d .i .1.

02:11 Now we want to know where this lens would focus an object here.

02:17 And so we can again use the thin lens equation...