An object is placed in front of a converging lens with a focal length of 20.0 cm . For each obje

step 1 This time we have a converging lens with a focal length of 20. So I just sketch a converging len

An object is placed in front of a converging lens with a...

Key Concepts

Converging Lens

Converging (or convex) lenses are optical devices that bend incoming parallel light rays so that they converge at a point known as the focal point. They are commonly used in various optical instruments for focusing light, and their behavior is governed by principles of refraction.

Lens Equation

The lens equation, given by 1/f = 1/d? + 1/d?, relates the focal length (f) of a lens to the object distance (d?) and the image distance (d?). This fundamental formula allows one to calculate the position of the image formed by the lens when the object distance and focal length are known.

Magnification

Magnification is a measure of how much larger or smaller the image is compared to the object. It is calculated by the formula m = -d?/d?, where the negative sign indicates that a real image formed by a converging lens is inverted relative to the object. Understanding magnification is critical for analyzing the size and orientation of images.

Real and Virtual Images

Real images are formed when light rays converge after passing through the lens, and they can be projected onto a screen, whereas virtual images are formed when the light rays appear to diverge from a point behind the lens and cannot be projected. The nature of the image—whether real or virtual—depends on the object's position relative to the focal point of the lens.

Transcript

00:01 This time we have a converging lens with a focal length of 20.

00:04 So i just sketch a converging lens like this, a principal axis, and the focal length is 20.

00:13 And a converging lens by definition has a positive focal length.

00:20 So i'll give it three significant figures as it has in the text.

00:24 So in the first case, part a, they have an image distance, i'm sorry, an object distance.

00:30 Of 40 centimeters.

00:33 So our object is out here.

00:36 It's about it is exactly twice the distance of the focal length.

00:40 So what kind of image will we get? well, because it's nice and easy, let's do a little ray diagram.

00:47 We can see what kind of that's supposed to go through the focal length there.

00:51 We'll see what kind of image we get.

00:57 Okay.

00:57 So the rays cross here, so we know it's a real image.

01:01 It's upside down or inverted and the size looks approximately the same but this is not exactly a two -scale diagram so now we can prove it with some numbers so using the lens equation one over f equals one over p plus one over q the focal length is 20 the image is what we don't know but the object distance is 40 so one over q so if we isolate q we have one over 20 minus one over 40 is equal to 1 over q and 1 over 20 can be written as 2 over 40 minus 1 over 40 and that's equal to 1 over q so 1 over q was 1 over 40 therefore q is equal to 40 centimeters so in this case the q is the same magnitude as the image distance i mean the image distance is the same magnitude as the object distance and the magnification therefore is negative q over p, magnification is going to be equal to negative 1.

02:09 Whoops, equals negative 1.

02:12 So what does that mean? we have an inverted image because it has a negative sign here.

02:18 It's equal to the size of the object because the magnification is 1.

02:23 It's not smaller or larger than 1.

02:26 And it is real because our object, sorry, our image is a positive.

02:34 So for the next one, they want us to do the same thing for an object distance of 10.

02:39 So i'm just going to quickly draw a new ray diagram.

02:43 In this case, the object is within the focal length.

02:48 So if this is our focal length here, that's 20, then our object is placed here.

02:54 So let's draw some rays.

02:56 This is a converging lens...