Zoom Lens. Consider the simple model of the zoom lens shown in Fig. 34.43a. The converging lens has focal length f1 = 12 cm, and the diverging lens has focal length f2 = -12 cm. The lenses are separated by 4 cm as shown in Fig. 34.43a. (a) For a distant object, where is the of the converging lens? (b) The of the converging lens serves as the object for the diverging lens. What is the object distance for the diverging lens? (c) Where is the final ? Compare your answer to Fig. 34.43a.

Video transcript

Hi, everyone in this practice problem, we're being asked to determine the location of S one prime, S two and S two prime where a student builds an optical system by placing coli two lenses, L one and L 25 centimeter apart. L one is a divergent lens with a focal length of F one equals negative 10 centimeter. And L two is a convergent lens with a focal length F two equals four centimeter. The student placed a matchstick at a distance of eight centimeter to the left of L one. And we're being asked to first determine the location of S one prime, which is the image produced by L one. Second. The image produced by L one is the object for L two. And we're being asked to determine the distance as two that separates the object from L two. And lastly third, to find the location of the final image of the two lenses of the optical system. The options listing the different locations and distances for S one prime S two and S two prime. So first, in order for us to tackle this problem, let's draw the system that we have in our problem statement. So in this case, we will have the first lens which is going to be a divergent lens with a focal length of negative 10 centimeters. So I'm gonna draw a divergent lens here and this is going to be our L one and we have L two on the right of L one, which is going to be a convergent lens. This is going to be L two and the distance between L one and L two is going to be five centimeter. The match is going to be placed eight centimeter away to the left of L one which is going to be indicated by this. And then I'm going to indicate that the distance is going to be eight centimeter away. So this is a system that we have. And then we want to recall that the object image relationship for a lens is given by one divided by S plus one, divided by S prime equals to one divided by F. So let's first look into the L one or the image produced with the divergent lens of L one. So I'm gonna indicate that with the first one. So we know that for L one F one will be equals to negative 10 centimeters given in the problem statement. And S one will equals to eight centimeter which is the distance between the match and the divergent lens. So therefore rearranging our object image relationship formula here we can get an equation for S one prime which going to actually equals to S one multiplied by F one, define it by S one minus F one. So let's substitute the values that we have S one is eight centimeter multiply that with negative 10 centimeter divide everything with S one which is eight centimeter minus negative then centimeter four F one calculating this, we will get S one prime to then be equals to negative 4. centimeter. So the image S one prime produced by L one is going to be virtual and position 44.4 centimeter away to the left of the lens L one. Awesome. So let's move on to the second part which is the image produced by L one or S one prime going to be the object for L two. And the distance S two that separates the object from L two is going to then be equal to S one prime plus the distance between L one and L two, which is going to be equals to five centimeter. So in this case, then S two will then be equal to uh 4.4 centimeter plus five centimeter. We are only looking at the absolute value because we know that uh the distance or the direction of S one prime is to the left of L one. So it's going to be 4.4 centimeter away to the left of L one plus the distance between L one and L two. Which is five centimeter. So that will give two to then be equal to 9.4 C meter just like. So also lastly, we are looking at the third part which is the image produced with the convergent lens L two or the final image we have the F two to be equal to four centimeter and S two to be equal to 9.4 centimeter which is obtained from the second part of the problem. The image uh distance as to prime produced by L two is obtained using the object image relationship for a lens again. And S two prime, similarly to S one prime is going to then be equal to S two multiplied by F two divided by S two minus F two rearranging or um substituting I mean substituting all of the values that we know we can then then get S two is 9.4 centimeter multiplied by F two which is four centimeter divided by S two, which is 9.4 centimeter minus four centimeter four F two. Calculating this, we can get S two prime to then be equal to seven centime meter. So the final image S two prime is going to be at C seven centimeter to the right of L two. So this is S two prime and then S one prime is a negative 4.4 centimeter or 4.4 centimeter to the left of L one and S two is going to be 9.4 centimeter or 9.4 centimeter to the left of L two. So there we have the answer to this uh problem and all of this uh answers will be summarized in option A where S one prime is going to be 4.4 centimeters to the left of L one, S two is 9.4 centimeters to the left of L two and S two prime is seven centimeter to the right of L one. So there we have it and there option A is then going to be the answer of this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please feel free to check out other lesson videos on similar topics available on our website. But other than that, that will be it for this one. Thank you.