A screen is placed a distance d = 41.0 cm to the right of a small object. At what two distances to the right of the object can a converging lens of focal length +6.55 cm be placed if the real image of the object is at the location of the screen? sf (Hint: Use the thin lens equation in the form s' = s-f so s'(s-f) = sf. Write s'=d-s and solve for s in the resulting quadratic equation.) shorter distance longer distance cm cm For this lens, what is the smallest value d can have and an image be formed on the screen?
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- Please ASAPAn object is placed 66 cm in front of a converging lens of focal length 9 cm. Another converging lens of focal length 14 cm is placed 26 cm behind the first lens. a) Find the position of the final image with respect to the second lens. cm b) Find the magnification of the final image.An object of height 5.25 cm is placed at a distance of 20.5 cm from the convex lens. Whose focal length is 12 cm. Then Calculate the following: (a) The position of the image in cm from the lens: Answer for part 1 (b) The magnification of the lens (Only magnitude): Answer for part 2 (c) Height of the image formed
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- An object is 6 cm in front of a converging lens with a focal length of 10cm. Draw a ray diagram (to scale with a ruler) to find the location of the image. Is the image upright or inverted, and Is the image real or virtual? Then I want to use the thin lens formula to find the image distance and the magnification. I got stuck in the middle of this problem and am confused. Thank you for the help!I continue to get the wrong answer. TIA.You place a magnifying glass of focal length 10 cm a distance of 8.5 cm away from a newspaper page. To someone looking through the magnifying glass, how large does the letter "a" on the page appear to be if its actual height on the page is 4.23 mm? Report your answer in mm. mm (+ 0.5 mm)