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Part AAn cbject / placed 35 cm to tne lelt of douibConvalengtn 40 cm. WhereInls objec: located?50 cm tne left ofthe Iens160 cm tne right nf the8 6 cmto Ihe right -2...

Question

Part AAn cbject / placed 35 cm to tne lelt of douibConvalengtn 40 cm. WhereInls objec: located?50 cm tne left ofthe Iens160 cm tne right nf the8 6 cmto Ihe right -280 cm lo Ine Iell ol lha lans60 cit thg leltol Ihe Iun8SubmliBeque LAnikerProvide FccdbackNext4m2go

Part A An cbject / placed 35 cm to tne lelt of douib Conva lengtn 40 cm. Where Inls objec: located? 50 cm tne left ofthe Iens 160 cm tne right nf the 8 6 cmto Ihe right - 280 cm lo Ine Iell ol lha lans 60 cit thg leltol Ihe Iun8 Submli Beque LAniker Provide Fccdback Next 4m2go



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A $5 \mathrm{~mm}$ high pin is placed at a distance of $15 \mathrm{~cm}$ from a convex lens of focal length $10 \mathrm{~cm} .$ A second lens of focal length $5 \mathrm{~cm}$ is placed $40 \mathrm{~cm}$ from the first lens and $55 \mathrm{~cm}$ from the pin. Find (a) the position of the final image, (b) its nature and (c) its size.

Okay. So for this question we are falling the format of previous at least 20 emotions. So in this we have to divide the given numbers and round off them to the provided significant digits. We have 26 divided by 10.1. The answer is 2.574-6 centimeter And we need to run them off up to three significant digits. So the answer will be 2.5 seven centimetre Production be we have 9.95 Single .15. The answer is 66.333. And uh two significant digits round off. So the answer will be 66cm options See? And we have 131.78 divided by 19.25. We have something with a square sending medium and the answer is well sending me the square can be reached and a centimeter multiplied with centimeter cancel this out 6.845 71 And in this question we can round off to four significant digits, so the answer will be 6.84 six centimeter square. We want to option D. We have 131.78 centimeter cube, Divided by 19.2 Centim. This God! This was too. The answer is 6.86354 sending me a square, And we have to run them about 23 significant digits. So the answer will be 6.8 six, sending his foot.

In this problem. You have two lenses, one convents when conch 8 50 centimeters apart. The complex has a focal in the 25 centimeters, and the con cave has a focal negative. 40 centimeters on were asked to calculate where the image is being produced. This object is two centimeters tall and the object is initially placed 30 centimeters in front the first lens. So to do this we can use the Finland's equation, which is one over d o plus one over D. I equals one of her out and we can solve for one over D I now we want to get d I by itself. So we're gonna go ahead, inverse, inverse the entire thing. And this gets us a distance of the image. One for the first limbs off. Ah, 150 senators. That means that it is somewhere over here. This is going to be labeled as 150 centimeters. And so if I take away the distance that is between these two, we're going to sell for a new distance of the object so distance of the object to as, ah, 100 centimeters and this is going to be a negative value because it is now on the right side of the lens and that we're going to repeat the same process. So d I equals one over one over s which is now negative 40 minus one over. And this gives us negative 100 and we get a new distance of the image as negative 66.667 centimeters. So that means that my image is being created to the left of the second blends and we can use that to determine our magnification Now are magnification is going to be de i one times d i to all over d 01 times d 02 So we get negative. 66.7 round up times 1 50 divided by 100 negative 100 times 30 gets the magnification about 3.33 So, like we said, this is going to the fact that this is a positive value. It means that my image will be up, right? So since the final image will be formed on the left of the convex lens, that is the same side is the original light bulb, which means that this final image will be virtual So this is a virtual image. Again, we got a distance of negative 66.7, which tells us, starting from our second lens are con que blintz that we are 60 about 67 centimetres in this direction. So my image will be located out here because that is beyond the 50 centimeters that separates the two lenses. So because the final image is to the left of my convex Maikon, Kate lends this is again a virtual image.

Hi. In the given problem, there are two lenses. The first one is a convex lands, and the second one is a concave alliance, and the school dances are kept at a distance off one at a distance of 50 centimeter between them, the focal length off first plants. The convex lens has been given us 25 centimeters, and the Oakland off second Convex lens has been given us minus 40 centimeters. An object has been kept in front. Off this, convicts lands at a distance off 30 centimeters. So as for sign convention for the first plans, this is minus 30 centimeter. In the first part of the problem, we have to find the final position off the image formed by the combination by this lens combination. Hence, for the first plans, use England's equation, which is a relation. Among the object is stance, image, distance and for Colette. So it says, one by Cuban minus one by even visible to one by F one. Or we can say one by Cuban minus one by minus 30 is equal to one by 25. Or this is one by Cuban is equal to one by 25 minus one by Kirby Hell, This Elsie, um, comes out to be 150. This is six minus five or one by 150. So we can say that image formed by this first plans. Really? Each one by this first plans should have seen at a distance. So 150 centimeters towards the right. So these are the incident lease he is coming from The object thes raise should have informed and image off this object a distance off 150 centimeter. But before that, this concave Elin's intercept the race So this concave lens will divert stories further away from that combination. Hence, for the second lands, this image will serve as the virtual object. So the distance for this virtual object comes out to be be to which is given by 1 50 minus 50. Or this is plus 100 centimeters for the second lands. And we know the focal length after his minus 40 centimeters. So using violence situation again one bite, que tu minus one by P two is equal to one by f two, or you can say one bike uto. This is missing minus one by 100 off. Positive sign equals toe one by minus 14. So this one like you too becomes one by 100 minus one by 14. Who's Elsie? Um, Comes out Toby 200. So this is Tu minus. Hi. What we can say this is minus tree by 200. So finally this Kyoto, the distance off image from by this concave lens comes out B minus 200 by cream or this is minus six 36.67 centimeter or roughly, we can say minus 67 centimeters from the second lens. So this is the answer for the first part of the problem and negative sign on Lee for sign Convention for the rial distance off, this final image is 60 70 67 centimeter from the second lands the works Right now, In the second part of the problem, we have to find the orientation off this image. So, using the expression for the next leader magnification, chemicals toe anyone into empty, or this is Cuban by even multiplied by Q 25 p two. So putting all these known values here this is 150 for Cuban minus 30 For Stephen. This is minus 67. 4. You so And for Peter Business 100. So ultimately it comes out to be positive. Hence we can say the final images erect. And as the final images area can be informed, toe the right off the lands combination. In the last part of the problem, we can see this final in each is virtual as only virtual images are red. Thank you.

Here we have been given that height of object at his statue is given as three cm. And the focal length of the lands have is equals 2 -6.2 cm. Let's move to party of this problem. So by by the formula of magnificence and weekend. That magnificence and equals two minus E. I. Whatever deal this is equal to at I divided at. Oh but from here we can rewrite this as he read it. He says B. i. equals two minus of H. I. That is 1.8 divided by actu. That is three multiplied where D. O. So Yeah I will be equal to -0.6 terms of video. Now let's use the lens formula we can write one divide. Ever focal length. F. Is equal to one day wherever image distance Plus one Day Wherever Object Distance. So let us substitute. So it will be one day wherever D. O. Plus the value of Here is -0.6 deal. So it is minus zero point six times of deal Plus one day whatever do you. But we can rewrite this as you can read her. This is One day where they were focal length is 6.2 cm. This is equals two one divided over. Do you multiplied with minus zero point 767 From here we will get the amateur of object distance of object U. is equal to 4.13 cm. So the lens would be placed at 4.13. Distance from the pin. No. Now let's move to part B. So here the distance of the image we have Calculated as -0.6 times of the distance of the object. So let us substitute the value. It will be -0.6. Early plant where you is 4.13 we have calculated So it will give us a value of -2 for it centimetres. Today I that his image distance will be -2.4 at centimeter The name is properties. Image is mrs erect. You may as well. We were actually also and it will be diminished. Humanist. Now let's move to part C. Of this problem. So we need to draw the diagram. So this is going to be there. Um Do you take them? That's right. The center line. So this is the center line and the object is at this distance. This is the object. No let's draw the diagram. Also the first reveal well here and it will reflect like this. The secondary will go like from the center of curvature. It will go like this third ray we'll go like this. We'll go like this from here. Then straight In the 4 3 the late at this point B the focus point, that one. So this is a phone. This is the focus point, everyone. Uh huh. When And this is point A. This point. This is and no from this focal length. This will come like this. So this will come like this, the it will close to this. And from here we will get the image at this point. This is the image. It means this is image. This is objective. A. B. Object. This is a dispute as Hope you like this.


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