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10(cm) distance Image -101.52.0Object distance (cm) (a) Converging lens...

Question

10(cm) distance Image -101.52.0Object distance (cm) (a) Converging lens

10 (cm) distance Image -10 1.5 2.0 Object distance (cm) (a) Converging lens



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A $1.0-\mathrm{cm}$ -tall object is $20 \mathrm{cm}$ in front of a converging lens that has a $10 \mathrm{cm}$ focal length. Use ray tracing to find the position and height of the image. To do this accurately, use a ruler or paper with a grid. Determine the image distance and image height by making measurements on your diagram.

All right, So we're gonna put this object or two lenses to get it to be perfectly displayed on a screen 110 centimeters away. So everything want to do is put it through the funds equation of our first diverging lens. So one of her F one is equal to one over s one plus one of her s one prime. We have one over. Negative 20 is equal to 1/20 plus one of her s one prime as one prime is equal to native 10 centimeters. And now we can look at the magnification for our first lens. M one is equal to negative s one prime over s one, which is negative. Negative. 10 over 20 which is huh? 1/2. Now, we also know that the final height is going to be negative, too. Centimeters, which is equal to m one m two h. We know M one is 1/2 ages ones therefore m two most people to negative four, which we know is also equal to negative s prime to over s two. So we can say that four s two is equal to s, uh to prime. Now, let's look at the lens equation for our second lens, one over F two is equal to one over. Well, the Dis Institute into one's is minus the image position of the first lens, which is, um esque one prime that plus one over. As to prime hold it won over. F two is equal to one over. Our distance is going to be well. We start with 100 10 centimeters. Means tracked 20 from the original opposition first lens. And we also need to subtract s to prime over here to get this distance T minus s one prime which is of course, negative. 10 centimeters plus one over S two prime. But we can express this expression has s too right so as to is equal to do you work this out? Ah, 100 minus as to prime so as to plus as to prise equal to 100. And now we have these two equations and we can plug him into each other and solve so well we want to express We want to solve press to prime so as to is equal to one of her fourth test to prime So one over fourth has to prime plus as to prime equals 100 has cheap prime equals 80 centimeters, which is what we want for part of our answer. Now you can return to our lends equation for a second lens one over F two equal to one over, Um, s two plus one over. As to prime plug in these numbers will, if s too with eight. Even as professed surprise Ativan as to is 20 centimeters. Right. So we get one over f twos, equal to 1/20 plus 1/80 and our focus of our second lenses and equal to 16 centimeters.

From till lens equation I can write one by F. Is equal to one by the not plus one by D. I. Simplifying it further. I can write the knot is equal to D. I. F by D. I minus F. So just putting the value in this expression I can't I did not is equal to 200 multiplication 20 by 200 minus 20 which is equal to 22 centimeter as our object distance did not Now. The lateral magnification formula is given by M. Is equal to minor. D. I buy dinner, simplifying it for this. Just putting the value I can like minus 200 by 22.22 which is the object value of the object distance. So on simplification I get the value of magnification at minus nine. So this is our final answer for this.

In order to solve this problem, we are going to use the well known thing lands formula. This Finland's formula says that one over the distance between the object and the lands, plus one over the distance between the image off the object and the lands equals one over the focal length according to the problem. Distance between the object and their lands. Musical toe 10 times the focal length. So what do you want? Easy kowtow. 10. Let's isolate D to from this equation. First, let's multiply both sides of the equation by 10 half. We see that then have gone so south and F cancels out as well. The creation becomes one plus, then have or are you too, he calls them. Let's move one toe the right side of the equation. Finally, let's isolate D to This is the final answer due to musical toe, then nines off the focal length

Hi. In the given problem, there is a contact lens whose focal length has been given us, then 30 centimeters and the object has been get in front off this convex lens at a distance off 10 centimeters. So, as per sign convention, it has taken as minus 10 centimeter. No, we will have to find the height off the image and the position off the image using a red racing when the height off the object was given, as each one is equal to 1.0 centimeters. So, first of all, in order to draw the rate racing, we have to draw its principal axis. So you consider this a scale in order to draw the re diagram? This is the principal axes. No, we consider optical center off the lens. Toby here, at this point, let me see optical center off this convex lens, then hide and death off this lens above and below. The principal axes are taken as 12345 five gaps above the principal axes and 123455 gaps below the principal axes. Hence this becomes a convex lens. No, we have to put the object at a distance off 10 centimeters. So we will count 10 gaps to the left off this convex lens. So 123456789 and 10. Hence, this is the position off the object and the object is having just Ah hi, doll. One centimeters. So this is the object a B in front of this convex lens. The focal length of this convex lenses 30 centimeters. Soto mark the focal length in both sides. Off this convex lens, we will have to count upto 30 gaps on both sides off this optical center. So up to here these are already 10 gaps means 10 centimeters Soap. 11 12 13 14 15 16 17 18 1920 21 22 23 24 25 26 27 28 29 30. So this is the second principle Focus, too, off this convex lens and similarly to the right hand side. One, 23456789 10 11 12 13 14 15 16 17 18 19 Plenty. 21 22 23 24 25 26 27 28 29 30 So this is the first principal focus off this convex lens for everyone. Now we will use the ray tracing, tow, obtain image off the subject. So first of all, you will consider array starting from point B off this object, this rate will move straight barrel to the principal, axes up to the midpoint off this convict lands, after which it will bend and we pass through the first principle focus off the convex lens. So here this is three sister scale. It should join these two times. We should keep it like this a little more definition. This a little down? Yeah. Finally. This is the correct orientation off the scale. Hence, due to the very small height off this object, we will have to face this problem while drawing redressing and separate dressing. So finally, this is the ray reflectively. It will pass through the first principal focus like this. Another legal passed through the hospital center and thes raise are diverging away from each other. It is clear. Yeah. So this is the read passing through the optical center. Finally, these rays are diverging away from each other. So on producing back, we will have to produce these rays backward. This is and then this one. So on producing back these rays after toe come from this point. Please raise Appear to meet at this point. So this is the image. Virtual image off this object. A dish be dish If we measure the height off the subject off this image this clear that the height off the image h two This is one complete block and a half lakhs. So this is 1.5 centimeter. The images upright means virtual. It is magnified and in the same side, off the lens where the object is No, we have to find the distance off this image. Find the image distance. We again for the scale here. But we need Noto. Use any scale because we are having the gaps. So we just count the gaps. So this is these are 10 upto here up to a B. Then this is 11 12 13 14. So there are total 14 gaps. Hence it represents. The image is at a distance off 14 centimeter. So the image distant skill comes out. Toby minus 14 centimeter and the height off the image is coming out to be 1.5 centimeter. Now we will compare these results in the second part of the problem with a mathematical results. So find the results. Mathematically, we will use lens equation. So using land situation, which says one by Q minus one by P, is equal to one by f. Their cue is the industry stance, which is on the scene. So this is one of the Q minus one by minus 10. It's going to for focal and this is 30. Hence one bite you. Is he going to one by 30 minus one by 10. Taking the Elysium becomes 30. This is one minus three or this is minus two by 30 which comes out to be minus one by 15. Hence, the image distance is minus 15 centimeter. While using the redressing the optimal the image distance Toby 14 centimeter hands. The result is nearly equal. Now we have to find the height off the image. Soto find the height off the image. We use expression for linear magnification, which says linear magnification is equal to the issue off image distance to the object that stands. And this is also Internet height of the image to the height off the objects, so plugging in on the values for Q disease minus 15 centimeters. Or be this waas minus 10 centimeter H two. We have to find each one watts 1.0 centimeters. So canceling all these things, the height off image comes out Toby plus 1.5 centimeters, which it's exactly as the found using very dressing. Hence the reserves obtained using the red raising are nearly equivalent to nearly similar toe. Those obtained mathematically thank you.


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