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Convex lens of focal length 1Scm produces virtual image whose size is thnce- the size of the object () Calculate the object and image distance. (2 Marks) () Calcula...

Question

Convex lens of focal length 1Scm produces virtual image whose size is thnce- the size of the object () Calculate the object and image distance. (2 Marks) () Calculate the height ol the object if the height of the image is 6cm. (1 Mark) (J) Explain the properties of the image by drawing the ray diagram; (2 Marks)

convex lens of focal length 1Scm produces virtual image whose size is thnce- the size of the object () Calculate the object and image distance. (2 Marks) () Calculate the height ol the object if the height of the image is 6cm. (1 Mark) (J) Explain the properties of the image by drawing the ray diagram; (2 Marks)



Answers

Consider an object of height $1 \mathrm{~cm}$ placed at a distance of $24 \mathrm{~cm}$ from a convex lens of focal length $15 \mathrm{~cm}$ (see Fig. 4.19). A concave lens of focal length $-20 \mathrm{~cm}$ is placed beyond the convex lens at a distance of $25 \mathrm{~cm}$. Draw the ray diagram and determine the position and size of the final image.

Solving party of this problem. So from his political middle equation I can like the expression like this here. The idea image distance efforts, focal length and do not eat object distance. So the image so the image for made age for virtually nature what you are in nature. Second it is upright and third it is reduced so this is the answer for part A. Now I'm going to solve part B. So in part B I can write the formula one by F. Is equal to one by denote plus one by D. I. Simplifying it further. I can write D. A. Is equal to do not have by the not minus L. Just putting the value here, I can write 1 20 minus 50 by 1 20 minus minus 50 on further simplification. I can write the value of D. I. Is equal to minus 1 20 multiplication 50 by 1 70 which is equal to minus 25.3 centimeters. The emergency at a distance of 35.3 centimeters behind the middle. That's why distances negative. So image form each so image formed here. Eat virtual. So the image formed here in this problem it what you want in nature. Now calculating the magnification of the middle so I can write the formula I am is equal to minor. D. I buy the note which is equal to minus minus 35.3 by 1 20. So finally I get em is equal to 0.3. Now the image height H. I. Is equal to a multiplication astronaut, which is equal to 0.3 astronaut. So the height of the image 0.3 times the height of the object.

In this problem who are asked to draw a re diagram for two convicts. Lanza's each having a focal length of four centimeters and a candles place tense in centimeters from the first plants. Now you are drawing this to scale. So it is important that you are using a ruler and measuring out all of these distances. Obviously, I can't do that. But I'm going to draw them out. And so again, you should be measuring out missing distances that I doing to scale. So what that means is this? Lenses are to 12 centimeters apart, and I am placing the focal off four centimeters. So about 1/3 of this distance again, I'm eyeballing it. You want to make sure that you're measuring this exactly to be four centimeters, and then our distance of the object is 10 centimeters away again. You measure this, I'm going to estimate it to be about here. This distance right here should measure distance of the object one as 10 centimeters. When you're drawing, you're a diagrams. The first rate goes parallel to the principal access, and then it goes through the focal now because this is a convex lens, the focal is going to also be a distance of four centimeters away, but on the other side, so again measured that out. So that's through the focal. And then the second ray diagram goes from the top of the object through the center of the lets. You can see where the two lines intersect is where my image is being created. So again, I'm representing my candle as a little error, which is because it's easier to drop now. For the 2nd 1 you are going to be doing the exact same thing. You're going to want to measure out that four centimeters again. I see the four centimeters is about here. One measure that number exactly. Measure it on both sides. And then same idea. Starting at the top of the object. We're gonna go parallel to the principal towards the centre, off the lens and then through the focal. And then these 2nd 1 is going to go through the center of the lunge, which is at the parallel, the lying down the center. So starting at the top of my object, you go through again. Where the two lines intersect is where my image is being created. You can see that the first lens did invert the image, and then the second lunch lens inverted the image back. Let's say that, actually forget.

Okay in this question we have to find out the images. Stands D. I. And the magnification. Yeah. To understand this question, we have to try the ray diagram of the lands. So we have given We have given the focal and S22 cm And the object height is given as four centimeter And the object distances given as 15 centimetre. So to get this question we have to draw the uh read a grandma. They can Excellent. This is a convex advance. This is the principal axis. This point is God Focus. And this is the other focus. So the object is placed between the focus and the optical center out there lens. So ray parallel to the principal access after reflection, they passed through the our focus. This is focus focus. So this point is gonna focus and ray through the optical center passed a state. And if we retrace back these two uh huh two lines then they will meet and uh point and the images from. So this image is an imaginary image and it is beyond do f It is beyond do if this is this is to F. So so from the middle formula we know that from the middle from latino they had one out of F. He went to one out of the I. bless one out of D not and one out of it is even is one out of 20 equal to one out of the eye plus one order 15. And run out of the eye, which is the major distances it went to minus one out of 20 two. He will do one out of 22 minus Run out of 15. And then this one of the of the eye, It went to -7 out of 330. So d I get equal to minus 47 point 14 centimetre. And the magnification is given us and magnification is u n s minus d i o daf d nutt. So the magnification can be written as- into -47 0.14 divide by 15. So we get, the unification is Plus 3.14.

For Problem thirty eight were given a converging lens with focal length of twelve centimeters, and we're told that a virtual image forms to the right of the lens at a distance of seventeen centimeters away. Since the images virtual, we know right away that the image distance it's going to be negative. So I read away, put a negative sign in front of the image distance. We're told that the height of the image, why prime to the millimeters and were asked to find the object distance and height which are leveled by S and Y respectively and were asked if the images director inverted if the object in the image on the same side of the lens. And also we're required to job array diagram. So in order to find us the object distance, let's use equation thirty four point one six. Okay, because we have enough information to determine s so as prime and f these air both known, they're given to us in in the in the statement of the problem and so we can very simply solved for s. So whatever s is one of her F minus one over as prime in that I could put these over a common denominator and then flip everything, just taking their reciprocal. And then I get house is equal to yes, prime divided by s prime minus No. Ask. Prime is negative. Seventeen centimeters. Half is twelve centimeters. So you knew where we have the product in the denominator, we have the difference between yes, and this leaves me with value s of seven centimeters. Okay, so that takes care of the object distance. Now we need to find the height of the object so we can use equation thirty four point one seven. So there we have any Barney magnification is equal to the object height, the image height, rather over the object type. But we don't know m. However, we do know why. Primed and we don't know why. So this isn't seem like it's enough to give me what I need. But there is a second part, uh, to this equation where the magnification is also equal to negative of the image Dessens over the object distance SMS primer both none. Therefore I can readily determined end once I know em. I can rearrange things in this equation to solve for why so from here. Hey, confine them. So remember, Ass is negative. Seventeen centimeters esperan. Sorry. And I guess we just found that seventeen years. And that gives me three point two five. Sorry. Thank you's me. Uh, not three point five in a urgent Teo. Also check my arithmetic has as I go along, you know, just plug in the numbers yourself. Sometimes I do make the stakes of these things. That is two points for three. And since Emma's positive, so am it's positive. That means that the image is erect. Okay, so we know M Now, we can use this portion of the equation if I rearrange a little bit. So if I move, if I multiply both sides by why and then divide by m, then I have y is equal to why primed over em? Why prime? They know my prime is eight millimeters and em. I know Emma's two point four three. Therefore, the object height is a three point two nine millimetres. Again, I will urge you to verify all the numbers for yourself. Now we'LL try the ray diagram on the next page. So here we have the optic access and they're converging lens. Um Let's say that's middle. And this is the focal length one side and on the other side. And, uh, from the problem, we found that the focal length is about twelve centimeters in the object was about seven hundred years. So somewhere midway come is the object. Oh, me pick a different car. So let's suppose that's our object. And now I will do is takes up some rays there. Going out? No. Yeah, uh, one that's going through the center like this. It's supposed to be a straight line and then another debts as if it's coming from this. This a focal focal point here. And then this Ray will go parallel TV uptick access. And if I extend that backwards, then I can draw my virtual image here. Mike, Settle. There couldn't, of course, be another Ray that is in fact, going parallel to your optic access. And then that will go through the to the other focal point. And that should interest like these lines but haven't drunk things to write. So I have trouble doing that one but the intersection of these two alliances and get me the virtual image. So this is yes, and this is that's playing and this sentence from hear, hear his F


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