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A photographic slide is to the left = of a lens. The lens projects an image of the slide onto wall 5 m to the right of the slide: The image is 83 times the size of ...

Question

A photographic slide is to the left = of a lens. The lens projects an image of the slide onto wall 5 m to the right of the slide: The image is 83 times the size of the slide: What is the focal length of the lens? (Give your answer in decimal with digits after the decimal point; using mm' (milli- meter) as unit:)diverging lens with focal length of -47 cm forms virtual image mm tall, 17 cm to the left of the lens_ Determine the size of the object: (Give your answer in decimal with digit after

A photographic slide is to the left = of a lens. The lens projects an image of the slide onto wall 5 m to the right of the slide: The image is 83 times the size of the slide: What is the focal length of the lens? (Give your answer in decimal with digits after the decimal point; using mm' (milli- meter) as unit:) diverging lens with focal length of -47 cm forms virtual image mm tall, 17 cm to the left of the lens_ Determine the size of the object: (Give your answer in decimal with digit after the decimal point; using mm" (milli-meter) as unit)



Answers

A slide projector has a converging lens whose focal length is 105.00 mm. (a) How far (in meters) from the lens must the screen be located if a slide is placed 108.00 mm from the lens? (b) If the slide measures 24.0 $\mathrm{mm} \times 36.0 \mathrm{mm}$ what are the dimensions (in mm) of its image?

Hi in this given problem size of the slide, it will treat as the object here for the given lens. And that size is given as 2.0 Centimetre width into 3.0 cm. The length focal land. Yeah, after the projection lens means the convex lens that is given us. Turkey, Centimetre plus 30 centimetre. And image distance means the distance of the screen instance of the screen will serve as the image distance. And that is given us then m. or this is 1000 centimetre. So using lens equation which is actually a relation among the object distance in this distance and or the length of the lens. This is one by S. I. Image distance, one by S. O. Object distances were too one by broken land. No, if we multiply both the sides by S. I. We will get first time side by side that will become one minus S. I by S. O. Is equal to S. I by F. Or finally we can save S. I by S. O. Which is actually the linear magnification also. So as I by S. O. That will be one minus as I by F. Or finally we can say linear magnification is F minus S. I. Invited by F. So finally plugging in all the known values here. This M will be given by 30 centimetre for focal lamp minus S. I. 1000 Centimetre divided by 30 again. So here it will come out to be minus 970 Divided by 30 cm by centimeter, canceling the centimeter. Finally this linear magnification here, I'm supposed to be approximately minus 32 Means the images 32 times in size as compared with the object. A negative sign is showing that image is in worked it. So Here this magnification 32 suggest that the dimensions the size of the object Will increase by 32. Therefore, size of image maybe 2.0 cm times 32 into 3.0 Centimetre times That is all to 32. So finally, the size of the image of the slide From south to be 64 centimetre, multiplied by 96 centimeter, which is the answer for the given problem here. Thank you.

Okay, so we're given the focal length as well. Aziz. The object distance and the object dimensions, which I called h a one and a show to on. Do we need to find the image, distance and the image dimensions? So we're going to start with the thin lens equation, which is, uh, one over equals one over. Ah, object distance Dio Uh oh, plus one over the image distance Whenever d I using that we can solve for the image distance. Um, which is one of F minus one over the O. Um, inverted. We can plug in our numbers from over here. Um, we find it's one over ah 100 instant the negative three meters converting for millimeters. Um, one of her. Ah, 103 times 10 to the negative third meters. And then all that inferred ID. Um, we solved that. Plug it into her calculator, and we find that the image distance is 3.43 meters, and that's our answer for part a. And then for part B, we're going to use the magnification equation. Um, which tells us, um, that magnification equals the image height over the object height, which equals negative Ah, image distance over object distance. Um, and for this problem, we only care about the dimensions. So we can take absolute value of that, Um, an unsolved for image height. And we find image. Height equals, ah, image, distance over object distance, times, object height. And early In this problem, height just means like dimension. So it could be either length or with which we have over here. So we're gonna find that each I one which I made with is equal toe d I, which we solve for up there. So 3.43 meters over D O, which is one of three times 10 to the negative third, sometimes each one, which is 24 meters. Um, 24 time millimeters. Sorry. 24 times 10 to the negative. Third meters, um, which is going to give us a chai one equals 0.7 99 meters and similarly for h I to plug in the same numbers for the distances. Um, and then this time, instead of h 01 we're gonna plug in h 02 which is 36 times 10 to the negative third meters. Um, and that gives us each I to equals Ah, 1.2 meters. And those are solutions for the image dimensions

In problem thirty were given. That object, which is a photographic slide, is the left of the lens. And it forms an image six meters to the right of the object itself. And the image height is given by if the image height actually eighty times the object height, but absolute value. Because we're not certain about the sign whether the image is director inverted, I am so given that the object is to the left of the lens, this part of the statement complies. That s is greater than zero. So the object distance itself is positive because it's on the incoming side of the lens. And from the statement to the left of the lands that averages six meters, I'm sorry that the images six meters to the left of the object that tells me that s plus as primed a six meters, six meters. Okay, so there's a little being important equation to use and the way to solve part ay to find the object distance will use this equation and additionally, evil use the equation for the magnification, which is as prime over s. So I got this equation from Division thirty four point one seven Where am I? Is equal to minus as primed over Yes, you know just took perhaps of value. So the minus sign goes away and Shuriken sold for s prime in terms of s so that I have s prime. Is it cold too? Mm Times s. And there's another part to this equation where I am is equal to that's put value here. Why prime over Why? But we know that Why Prime is anyway. So if I put anything why over why? That's just a so I have that as prime as he called too. Anything? Yes. If I substitute this result into this equation for as prime than I get let me write that here. That s plus a yes Is it called you six meters or as times of one plus eighty which is eighty one is equal to six meters and then I just divide both sides by one. Don't tell me that as is equal to zero point zero seven four meters and right away I could see that if I take away the the absolute value that cast plus s primed is positive number. It's six meters. So ask Prime It's office to be positive in order to give me if ask zero point your sound for you know to give me that Not six meters so yourself. Party a That's soft Party is thie image erect are inverted. So now I have I know that s plus as primed his eagle team six meters. So from here I can determine that as primed itself is Palmolive and another way too to figure that out kind of reason that the that the lens has to be between the the object and the image for the image to form in the way of that, huh? So now determined whether or not the images Erector inverted. So let's use mm is equal to minus X primed over s and we know that s is created and zeros as it's positive. That's what me and my ass is great and zero as prime did also positive. So if I have to positive numbers here, so ask Prime over asked will also be positive. But then I have a minus sign in front so than him is negative. And if m is negative, that implies that the image is inverted. Okay, and let's do part. See where we asked Find focal length So the focal length is one over We use this equation will never has plus one ever ask Primed is equal to one over have. But you may say OK, we know as but we don't know ask prime and we don't know F But actually we can quite easily find s prime from this equation since they already know ass So then we can have as prime in and the only unknown will be out. So let's actually no write that out as prime is equal to six meters minus s And that's six meters minus zero point zero seven four meters which is, uh, gives me an s as primed of five point nine three meters. Okay, so now I can, um just off this equation for absence and no Ason s prime. So f is equal to as primed as over. Yes, prime plus s, which is I see your point zero seven four meters times five point nine three meters divided by Well, this had better be six meters and that gives me that efforts zero point. See? You know, seven three eaters note that after is positive. So let's, uh, do party f is positive. Therefore, a happy converging men's

But a off the human problem were asked to find their distance to the image. We can write the distant that image from the Formula One over effort. Tickle toe one or D I. Plus one over deal. Then we can find a d. I hear and substitute the values, which is 100 times 10 to the power minus three meters minus one, divided by one or three times dented power minus three meters whole minus one. Simply find this we get here. 3.43 meter is our distance to the image point. What me? We can write the relationship off height of the image divide by. Out of the object is a cool two more off the minus the r a distance the image you ever distance to object. Then we can find the dimensional image. So damage in one is a child. One that is a cool too, are the eye Which one times d'oh! So some street in while you see a d. I V Here it's leaf for 43 times 0.24 divided by gov. Is there a point to one old tree? So we find this, we get there a lost dimension to be 0.799 readers for a second dimension chai to physical to D I h 02 divided by deal. So this is 3.43 times. Here we have 0.36 divided by d'oh which is their 0.1 over three. So we find this. We get edge I to to being 1.2 meters. So this and these this is the dimension off the image.


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