An object is moving toward a converging lens of focal length ƒ...

Question

An object is moving toward a converging lens of focal length ƒ with constant speed v_o, but its distance d_o from the lens is always greater than f. (a) Determine the velocity vᵢ of the image as a function of d_o. (b) Which direction (toward or away from the lens) does the image move? (c) For what d_o does the image’s speed equal the object’s speed?

Accepted Answer

Welcome back. Everyone. In this problem, a convex lens is approached by a small insect flying toward it with a steady velocity you the lens which has a focal length F is positioned such that the insect is at a distance from it always greater than F. An image of the insect is formed by the lens at a distance si from the lens first express the speed U I of the image in terms of the insects, distances from the lens. Second determine whether the image moves toward or away from the lens as the insect comes closer. And third at what distance. So is the speed of the image equal to the speed of the insect? For our answer choices? A says UIE equals F squared U divided by minus F all squared, the image moves towards the lens and so equals three F. If the speed of the image is equal to the speed of the insect, B says U I equals F squared U divided by so minus fr squared, the image moves away from the lens and so equals two FC says U I equals F divided by so minus F, the image moves away from the lens. And so equals two F and the D says U I equals F divided by minus F, the image moves towards the lens and so equals F. Now, let's start with the first part of our problem. OK. We want to express the speed of the image in terms of the insects distance from the lens. Now, the first question we have to ask ourselves is what do we know about the speed of the image that's going to be related to the distance? In each case, whether it's the object distance, the image distance, uh the actual speed and the image speed. Well recall OK, that speed OK is actually the derivative of distance E with respect to time. So that means if we have some equations for or distance or object distance, so and or image distance, si then we should be able to differentiate and solve eventually for the speed of the image in terms of the insects object distance. OK. Now what do we know about the object and image distance? Well, we also know by the lens formula that the reciprocal of the focal length one divided by F equals the reciprocal of the object distance plus the reciprocal of the image distance. OK. Now, if we were to solve here for our image distance, OK. Uh So for our image distance, OK, then we also know that the reciprocal of the image distance equals the reciprocal of the focal length minus the reciprocal of the object distance, we combine both equations on both terms on the right hand side to get the object distance minus F divided by FSO. And thus, we can find the image distance by reciprocating which will get it to be equal to FSO divided by. So minus F now that we have an expression for the image distance, we can go ahead and solve eventually for the image speed. OK. Since we know that it's the derivative right? Then for our object distance, since the insect is moving towards the lens with a constant velocity, U then U the insect's actual speed is going to be equal to the negative value of the derivative of the object distance. OK. With respect to time where the negative sign indicates that the object distance is decreasing. Now let's differentiate the image distance to find the image velocity. Now we know OK that the derivative sorry the image distance then the image velocity sorry is going to be the derivative with respect to time uh for image distance. And we know that the image distance is equal to fso divided by the object distance minus F. So now how can we go ahead and differentiate this term? Well, we can differentiate by using the caution to OK. Because by the quotient rule, it tells us that if we have a function. Yeah, or rather let me write it this way. Let me write it in another way here to make it a little simpler if we're differentiating a quotient, OK. X divided by Y, OK. Then the derivative is going to be equal to X prime Y minus Y prime X divided by Y squared where X prime and Y prime represent the derivatives of X and Y respectively. OK. So we can apply this rule to go ahead and differentiate our image distance. OK? Because now if we let X OK, we let X be equal to our numerator here which is FSO OK, then that means its derivative is going to be equal to write our constant F multiplied by the derivative of our object distance. And now if we think about it, we know that U is the negative value of the object distances derivative. So thus the derivative of the object distance is going to be equal to negative U. Thus, the derivative of X is negative fu likewise, if we take our denominator which is our object distance minus F OK, then it's derivative is going to be equal to the derivative here of so minus F, the derivative of our constant F is going to be equal to zero. So that's also the derivative of so with respect to time which is going to be negative U know that we have those values. Thus, we can differentiate using the quotient rule which then tells us that our image distance is going to be equal to negative F UK X prime based on a quotient rule multiplied by minus F minus FSO multiplied by negative U. And now all of this is going to be divided by S or minus Fr squared. Notice here we can factor out fu from our numerator. OK. So that's gonna be fu multiplied by negative minus FK plus. Yeah. And all of that is being divided by S minus Fr squared. Now negative plus those two cancel and F you multiplied by a positive F is gonna be negative negative F. So that's positive F that's gonna give us F squared U. So that's gonna be F squared U divided by so minus F all squared. So what does this mean? Well, this just tells us then that the speed of the image in terms of the insects distance is F squared U divided by the E but divided by the object distance minus F all squared. We've solved the first part of our problem. Next, we need to determine whether the image moves toward or away from the lens. Let's look at our expression notice here that our image velocity is positive because it only involves positive quantities. So that means the image is moving in the same direction as the insect. However, since the image is on the opposite side of the lens from the insect, this means the image is moving away from the lens as the insect moves closer. OK. And now for the third part of our problem, we want to find when the image speed equals the insect speed. Now, in this case, that means you I is going to be equal to you, which means that F squared U divided by so minus F all squared is going to be equal to U. Now we can cancel U from both terms, we can square root both sides to get rid of the square. OK? And no, when after we do that and cross multiply, then we should get our object distance here. Our object distances minus F equal to F OK. Which tells us then that our object distance equals two F. So when our image, this image speed equals the insect speed, the object distance equals two multiplied by the focal length. Thus, when we look at our results and compare it to our answer choices that tells us that B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Key Concepts

Lens Formula

Magnification

Velocity of the Image

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