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50. (2 points) A diverging lens forms an image 1/3 the size of an object that is 30 cm from the lens (a) What is the focal length of the lens? (b) Are the object an...

Question

50. (2 points) A diverging lens forms an image 1/3 the size of an object that is 30 cm from the lens (a) What is the focal length of the lens? (b) Are the object and image on the same side of the lens or the opposite side? (c) What is the power of the lens?

50. (2 points) A diverging lens forms an image 1/3 the size of an object that is 30 cm from the lens (a) What is the focal length of the lens? (b) Are the object and image on the same side of the lens or the opposite side? (c) What is the power of the lens?



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A 3.0 -cm-tall object is placed $15.0 \mathrm{cm}$ in front of a converging lens. A real image is formed $10.0 \mathrm{cm}$ from the lens. a. What is the focal length of the lens? b. If the original lens is replaced with a lens having twice the focal length, what are the image position, size, and orientation?

Everyone here it is, given our diverge. England's or focal length of 25 centimeters. Since it is divergence of focal lenses, negative object is placed 18 centimeter toe the left. We have toe fire audition off image. Okay, I'm okay. The space off image and height during the lunch followed up substituting the web. So a ministry state you will get right. Since what a major states we is negative. That is, It is move up left. So, images to the left off the lens. No. Yeah, flips on Dispense off image is, um 10.46 centimeter. Uh huh. Now, magnification will be be upon you. That is minus 10.46 upon minus 18 to that is quite private, since magnification is less than one. So hide off a mate is less than high dog object. There's all times for about

Because this is the diverging thence. Focal length is negative. 15 centimeter image distance is given by five centimeter, so we will use the lance form. The one over ethic was one over object distance plus one over image distance. From here, we would find object. Distance equals one of the F minus one over the I in his distance in verse off that that comes out to be negative. Whether were 15 negative 1/5 in divorce in verse it was that made it can come outside. 1/15 just 3/15. It was negative. 15/4. Sending me did so. Object position Deal equals negative. 15/4 is 3.75 centimeter, which means this is ah, right to an object. Now magnification is given by immense distance over object distance absolute value that is five over 15/4. So that comes out to be 4/3. So magnification is also expressed as image height equals magnification, times object hide. So object Hide is image height divided by magnification. So that comes out to be too divided by the magnification. Becomes 3/4. So that is 1.5 centimeter then. Now we are changing the focus event to last 15. So we are starting part B. Now we're changing the focal lead. Two plus 15. So we had tried to find the image distance right now, and images distance will be one over F minus one of our d, all in verse of that which is 1/15. Thus once over 3.75 in verse and I'm going to use the calculator right now. I give up. Ah, 50 inverse plus 3.75 inverse, That comes out to be surprisingly nice. Three centimeter. So that is the image position now. And this is ah, really image. Really major three centimeter magnification comes out to be immense distance over object distance. So 3/3 0.75 Using that e mais height will come out. Toby magnification dives object height, which is 1.5 Sandy Mido We Jeez. 4.5 divided by 3.75 equals 1.2 centimeter

So we have two lenses we have. Let's l one and then we haven't l ttle and sell to the first lens. L one is a diverge England's and its focal length is negative. 10 centimeters. The second land plants is a converging lens and the focal length is 30 centimeters and the two lenses are 20 centimeters apart and an object is pleased. 10 centimeters from L one to Tokyo is 10 centimeters. So we have to find the final image. Now for finding the final image, we can first find the image form, but one and then the image becomes the object for l two. So we start with one over Dio plus one over G. I is equal to one over f and this would be one or 10 plus one over G. I is equal to negative 1/10 or one over the I is equal to negative 1/10 minus one over can. So d I is equal to negative five centimeters. So this means that this image is formed to the left off l one So the images form somewhere around here. So this is the image. Now we're gonna take this image and make it the object for l two. So now, for l two, we would say D four is now equal to 20 minus negative fight which is our 25 centimeters. So we say one hour Dio plus one over G I. Is he going to one or F 1/25 plus one over d. I is equal to one over 30. So what, over the I is equal to 1/30 minus 1/25 solving this gives me d I is equal to 150 centimeters a d. I is equal to negative 1 50 centimeters. So that means that the image is located a distance 150 centimeters left off l two. So that's Ah, part one. Now let's go to part beat. So for part B, since there are two lenses so the overall magnification will people to m one times m two and m one will be negative. D i over the old for the first lens times negative t I were the old for the second lens and now the d I for the second lands for the first lens was negative finds its negative off negative 5/10 times the negative off negative, 1 50 over 25. So the overall magnification comes out to be able to see, and now we know that each eye over each hole is equal to the magnification. So teach I over three is equal to three, and that gives the child is equal to nine points year old centimeters.

Solving party of this problem. So here we have given a physical to 20 centimeters so changing it in meters, I can write 0.2 little meter so we know that power is equal to one by, yep. So just putting the value of F. Year. Finally I get the answer at five diabolical. Now solving part of the of this problem. So here F is given edge minus 50 centimeter, so changing it in meter, I can like the expression age minus 500.50 m. Now we all know that power is equal to one by F. So just putting the value of F. Year, so minus point 50 meet that Which is equal to -2 directed as our final answer for part B. I hope you understand how I solve both the part. I just use this simple formula that is power physical to reciprocal of the focal length. That's why I like that to get the answer.


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