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4. Three thin lenses are aligned along their optical axis. The first lens has 8 +24fcom focal length: The second lens has a ~18.0-cm focal length and is placed 18 ...

Question

4. Three thin lenses are aligned along their optical axis. The first lens has 8 +24fcom focal length: The second lens has a ~18.0-cm focal length and is placed 18 cm from the first lens. The third lens has a +12.0-cm focal length and is placed 9.0 cm beyond the Locateete position of the image of an object located 36 cm from the first second lens lens. For simplicity assume light is propagating from left to right:

4. Three thin lenses are aligned along their optical axis. The first lens has 8 +24fcom focal length: The second lens has a ~18.0-cm focal length and is placed 18 cm from the first lens. The third lens has a +12.0-cm focal length and is placed 9.0 cm beyond the Locateete position of the image of an object located 36 cm from the first second lens lens. For simplicity assume light is propagating from left to right:



Answers

An object of height 3 cm is placed at a distance of 25 $\mathrm{cm}$ in front of a converging lens of focal length $20 \mathrm{cm},$ to be referred to as the first lens. Behind the lens there is another converging lens of focal length $20 \mathrm{cm}$ placed $10 \mathrm{cm}$ from the first lens. There is a concave mirror of focal length 15 $\mathrm{cm}$ placed $50 \mathrm{cm}$ from the second lens. Find the location, orientation, and size of the final image.

Okay, so we're doing Chapter 34. Problem. 95 year says three lenses, each with a focal length of 40 centimeters. Our line on a common acts CeCe adjacent lunges air separated by 52 centimeters and find the position of a small object on the access 80 centimeters from the first wins. Okay, so we know that we're gonna have to find the image of each lens deceptive first and serious here, and use each m ej of the lens as the virtual object for the next. So for Linz one we know that s equals 80. So s prime equals 80 times 40 over 80 minus 40. This becomes 87 years. So it s crime for the 1st 1 is 80 centimeters to the right of the first lens. That means for Linz to the object. Distance now becomes the becomes 52 minus. Hey, I'm sorry. That is wrong. Let me draw this out because I was not doing it right here. So he was first lens second winds, third buttons, right. 52 here. 52 here. And the object here Waas 80 away. So if we become 80 away on the other side that means this being a D that s two becomes negative a D minus 52 or negative 28 centimeters because it's negative because from the right side of the second winds, so using that we consult for us prime of two, this becomes negative 28 times, 40 over negative, 28 minus 40. So that is a positive 16.47 So if we're 16.4 centimeters to the right of the second lens 16.4, that means for Lin's three. We haven't object distance that is 52 minus 16.4. And it's positive because it's to the left of the third linds. So this is 35.53 centimeters. So S3 image is 35.53 times 40 over 35.53 minus 40 and this becomes negative 318 centimeters. So the final image is 318 centimeters to the left of the third linens. Or we can also rearrange that you made three birth 3 18 minus 80 minus 52 minus 52 and not equals 134 centimeters left of the original object. Awesome

Eyeglass. He's 10 lens and it is like that on you. Look at it from this direction or that direction. It appears to be similar, however, depending on the design. If you look at it from this side, the distortions ah, minimum.

To the chapter lens and optical system and it has two lenses that are used to create enemies for the source of height. It's not equals to 10 centimeter tall and it is located at a distance did not one equals to 30 centimeter to the left of the first lines as shown in the figure and Lebanese by convicts. Concave lens made up of crown grass of refractive index N equals to one point double five and radius of curvature. R one equals two. R. Two. This is equals to 20 centimeters. Okay, so uh we can see here that are one will be negative. So are one can be written as minus 20 centimeter and are too can be written as plus 20 centimeters from the diagram. Okay. Next we have that. We have the length L two is 40 centimeter to the right of the first lens. So we can say that distance d between the lenses is 40 centimeters. Okay. And the length L two has focal length have two equals to 30 centimeter which is converging lens so we have to determine the location of the final image formed. Okay. Position of the finally makes formed so related to the object. Okay, so now for the first lines we can apply lens makers formula so we can write that one by D. O. Plus one by day. I it is equal to n minus one of one by our one minus one by. Are you okay? So for the first lines weekend right that day I won this will be equals two day I won. This will be equals two 1.55 minus one. We can substitute the values here so one by D. 01 which is equal to one by D. 01 which is equal to 30 centimeters. So 30 plus one by D. I. One that is equal to 1.55 minus one. And R. One which is equal to 20 centimeter minus 20 centimeter and our two is 20 centimeters. So from here after solving we get the Ivan that is a major distance from the first planes. That is equal to minus 11.32 centimeters. So minus is saying that the M. S. Is towards the left. Okay so it means the object virtual object for the second lines. It will be at a distance D. 02 That will be equal to D. I. One plus distance D. Okay so this will be equal to 11.32 centimeter plus D. Which is 40 centimeter. So it means it comes out to be 51.32 centimeters towards the left. So that is in the negative sign. Okay so we can make this negative sign. Okay so uh the object from the second line is located at this distance. Now applying the same lines makers formula for the line situation for the second lines because we have focal length here so we can write that one by D. I. Two. That is equal to one by F. Two minus one by D. D. Or two. So substituting values. So after that is equal to 30 centimetre minus one by D. 02 which is obtained as 51.32 centimeters. So this becomes one by D. I. Two. So from here after solving we get A. D. I. Two, That is equal to 72.2 cm for towards right of the towards the right of the lens. To that is help you. Okay so we have to calculate the in this position from the object. So the amis position from the object. This can be calculated as a. D. So it will be equal to 30 centimeters. That is objected exchange from the length first and the 40 centimeter, which is the distance between Elvin and L. Two and two which is towards the right, which is equal to 72.2 centimeter. So the distance of the maze finally maze from the object comes out to be 142 centimeter. Okay, so this becomes the answer for this question. Okay? Or we can clearly right that it will be 1 42.2 centimeter, which can be nearly taken as 1 42 centimetres. Okay, so this becomes the answer for this question. This is the location of the final miss from that object and next we have to determine this position by sketching ray diagram. So the radiogram can be drawn like this. Ok, so this will be the radiogram for this situation in which this is length one and this is the lengths to which have a total distance between them as a D. Okay, and this is the object which is pledged here. So this length one is making an M. Is here, Which is in the left side. So that's where that distance was negative. So this is the distance of objective one. And this is the distance. Sorry, this is object and this is a mess. Okay, This is the distance of the M. S. Okay, now we can calculate this objective distance. That is D. 02 which is D. I one plus this distance. D. So D. 02 that become D. I one plus D. Here as we've done in the calculations. Okay, so now this D 02 will form an M. S. D. I to adhere. So this will be the final mm. This is our final aim is from the this lens, converging lens. Help. So this is the radiogram for this question. Okay, These are the radius of curvature for the diverging lens. Okay?

Here is all the information we need. I was first start by finding the image position through the first lens. So first we start the equation. One over f F one is equal to one over s one plus one of her s prime one which equals 1/5 centimeters people to on over four centimeters, plus one over s prime one as prime one. He was negative. 20 centimeters. Now we could do this. Second lens one over F two is equal to one over the difference between the distance between the two lenses and the emits position from lens one. So T minus s prime one where the image from lens one become the object of lens too. But we need to find the position relative to ones, too. So that's what I'm doing there. And plus one over s prime too. Plug in our numbers and we get one over. Negative. Eight centimeters is equal to one over 12 centimeters. Minus and negative. 20 centimeters plus one over S two. Prime has to prime cools negative six point four centimeters. So that's gonna be our image. Finally, my position. Six. My four centimeters to the left. of Lens too for coming in initially from the left. Now let's look at magnification. Saman cation for our first lens is equal to the inverse of the first image position over the first object position which is just negative. Negative 20 who were four centimeters. So that's gonna be five him to the same thing that is going to be negative s to prime over S two, which is Nick. It is negative. 6.4 over again. 12 minus negative 20 which is 32 president of 0.2. We know that our final height is going to be, and one and two times H, which is five times Syria point to tens are socialite one centimeters or final height is equal to one centimeter and because hopes because it is positive image is non converted.


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