Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
The distance of the image formed is 44.27 cm and the magnification of the image is 0.7.
Location of the image
The image distance formed by the lens is calculated as follows;
1/v + 1/u = -1/f
where;
- v is the image distance
- u is the object distance
- f is the focal length of the lens
1/v = -1/f - 1/u
1/v = -(-1/26) - 1/63
1/v = 1/26 - 1/63
1/v = 0.022588
v = 1/0.022588
v = 44.27 cm
What is magnification of lens?
The magnification of a lens is defined as the ratio of the height of an image to the height of an object.
Magnification of the image formed
The magnification of the image is calculated as follows;
Magnification = image distance/object distance
M = 44.27/63
M = 0.7
Thus, the distance of the image formed is 44.27 cm and the magnification of the image is 0.7.
Learn more about magnification here: https://brainly.com/question/1599771
#SPJ1
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.