It is desired to magnify reading material by a factor of 3.0 x...

Question

It is desired to magnify reading material by a factor of 3.0 x when a book is placed 9.0 cm behind a lens.

(a) Draw a ray diagram and describe the type of image this would be.

(b) What type of lens is needed?

(c) What is the power of the lens in diopters?

Accepted Answer

Welcome back, everyone. In this problem, an optical lens produces a magnification of three when an object is placed 12 centimeters away from its lens, determine the type of lens required to produce this magnification. Describe the characteristics of the image formed real or virtual operate or inverted and draw a ray diagram illustrating the image formation and then find the lens power in diopters. And here we have a diagram, a graph where we're going to sketch our lens and the ray diagram. Let's start with the first part of the problem. What type of lens is required to produce this magnification? Well, if we go back to our problem statement, we notice that our magnification of positive three means that the lens must be converging because the image is magnified and the diverging lens do not produce a magnified image. So on our diagram already, we can put our converging lens. OK. So let's place our converging lens in the box. Now, in the next part of our problem, we want to describe the characteristics of the image formed and draw a ray diagram illustrating the image formation. OK. So that means on our diagram, we will need to put our virtual focus, our focus and where the image will be formed. OK. Now, if we think about it based on the information that we have the magnification is positive, the image is magnified. Then that tells us the image is going to be virtual. Here, we have our image or sorry, our object which is placed 12 centimeters away from the lens. And if we look at the units on the x axis of our graph, OK. We can tell that each unit is three centimeters. OK. In other words here, if this is the center, then this would be, well, you know what, let me just let me just put that in here. OK. So with our center here, OK. And this should probably be, I should probably move this over a little bit. Can't make it a little bigger, then we can tell OK that each unit is three centimeters. So if this starts at zero and this is 369 and then 12 centimeters OK. That's where our object is. So now the question is how are we going to figure out first of all where our image is going to fall the image distance? Well, we can use the magnification formula. OK. Recall that the magnification is going to be equal to the negative value of the image distance divided by the object distance. We know. Therefore, that means the image distance is going to be equal to negative MD O where as we said, the magnification is positive three. So that's gonna be negative three multiplied by our object that stands of 12 centimeters, that gives us an image distance of negative 36 centimeters. And that makes sense because the object must be between the virtual focus F prime and the lens. OK. So this confirms that the image is virtual and that it forms on the same side of the lens as the object. So now if this is at 12 centimeters, then if we go up to negative, if we go up to 36 OK. On this side, 1518, 21 24 27 3033 and then 36. OK. So our 36 centimeter mark is here, then that will help us to identify the point where rays converge. OK? Because now if we think about it here from our object, it goes towards the lens, OK? And no, we know that for this one, it's also going to go right at the central point here. OK. Right. That the essential point and if we extend that upwards in the other direction, well, let me kind of draw it on it just to make sure that we get the, get it right. OK. Then no, it makes sense here that our virtual operate and enlarged image will be at this point. OK. So now we just draw an arrow here to it. OK? And you know what I'm gonna shade this, I in red, then this is where we have our image distance, OK? And no, if we think about it here, from this point, if we put it where they converge, OK, then now we know that we can continue our line here again. Let me just make sure that it's uh make sure that it's straight through. OK. And no, at this point is where we will have F OK. And we know that we have our rays going in this direction. OK. Now, as we said, the virtual focus F prime must be after the object because the object is between F prime and the lens, we can use the lens formula to calculate the focal length F OK. Now, by the lens formula, we also know it says the reciprocal of the focal length equals the reciprocal of the object distance. In this case, 12 centimeters or 0.12 m plus the reciprocal of the image distance, negative zero point or one divided by negative 0.36 m. Now, when we go ahead and simplify here, we get this to be 59th per meter. OK? And now that means our focal length, if we reciprocate both sides is going to be equal to 9 m divided by 50 which is equal to 0.18 m. OK. So now when we go back to our diagram, we know that it's gonna be right here. The virtual focus F prime is gonna be at our 18 centimeter mark. OK. No, that we've completed the second part of our problem. We know that the image is virtual. OK. And we also know that it's upright and we know that uh we also know that it is an enlarged image right next, we need to figure out the lens power in diopters. Now, how can we find that? Well, finally, we also know that the lens power in the TERS is going to be the reciprocal of the focal length. So that's going to be 50 divided by nine which equals 5.56 diopters or to two significant figures, approximately 5.6 diopters. Therefore, this is a converging lens. OK. The image, sorry this let me go back to the problem. The characteristics of the image is that it is virtual upright and an enlarged image and the lens power in diopters is 5.6 diopters. Thanks a lot for watching everyone. I hope this video helped.

It is desired to magnify reading material by a factor of 3.0 x when a book is placed 9.0 cm behind a lens.
(a) Draw a ray diagram and describe the type of image this would be.
(b) What type of lens is needed?
(c) What is the power of the lens in diopters?

Key Concepts

Magnification

Lens Types

Lens Power

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