If you put an eraser 28.0 cm in front of a lens, an image of the eraser is formed 42.0 cm behind the lens. (a) Find the focal length of the lens. Is the lens converging or diverging? (b) What is the lateral magnification of the image? (c) Is the image real or virtual? Erect or inverted? Larger or smaller than the eraser?
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If you put an eraser 28.0 cm in front of a lens, an image of
the eraser is formed 42.0 cm behind the lens. (a) Find the focal length
of the lens. Is the lens converging or diverging? (b) What is the lateral
magnification of the image? (c) Is the image real or virtual? Erect or
inverted? Larger or smaller than the eraser?
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- On one side of a diverging lens of focal length 42.7 cm, you position an object of height 3.27 cm somewhere along the principal axis. The resultant image has a height of 1.31 cm. How far from the lens is the object located? 51.1 cm 63.9 cm O 102.2 cm O 89.5 cmCan you help me calculate the height. I kept getting it wrong.Suppose you have a concave lens of focal length f 30.0 cm. An object of height 5.5 cm is placed 35.0 cm from the lens. Using the algebraic method, determine: (a)the image distance; (b) the image height; (c) the image magnification; (d) whether the image is real or virtual; (e) whether the image is upright or inverted.
- 20. An object is 25 cm to the left of a diverging len. A converging lens (focal length = 12 cm) to the right of the diverging lens as shown in the figure. The final image from this lens system is at 17 cm to the right of the converging lens. (a) What is the focal length of the diverging lens? (b) If the object is 10 cm tall. What is the height of the image? (c) Is the final image upright or inverted? -25.0 cm- 30.0 cm 17.0 cm-(a) A 4.0 cm height object is placed 13.0 cm from a concave mirror whose radius of curvature is 12.0 cm. Determine the image distance and image height. image distance = cm image height= cm (b) A thin diverging lens with focal length of 16.0 cm is used in APEX project. You want to use this lens to form an upright virtual image that is one-third (1/3) of the height of the object. Where should the object be placed? object position = cmYou place an object 21.2 cm in front of a diverging lens which has a focal length with a magnitude of 12.2 cm. Determine how far in front of the lens the object should be placed in order to produce an image that is reduced by a factor of 2.70.cm
- 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 formedAn object that is 6.00 cm tall is placed 4.00 cm from the surface of the diverging lens.The magnitude of the focal length is 20.0 cm. a)Find the image position and the size b)Is the image real or virtual? Justify. (c)Is the image upright or inverted? Justify(a) As shown in Fig. la, the object and the screen is separated by 1 = 100cm. When we put a convex lens in the middle point between the object and the screen, a clear image is formed on the screen. What is the focal length of the lens? You may use either ABCD matrix or other methods. 100cm Fig. la (b) Now we replace the lens in (a) with an unknown lens (focal length f), as shown in Fig. 1b. Assume the distance between the object and the lens is S cm, what is the ABCD matrix from the object to the screen? What can we know about this matrix if a clear image is formed on the screen? 100cm P1 20cm P2 Fig. 1b (c) When we move the lens back and forth, we find that there exist two positions (Pl and P2 as shown in Fig. 1b) where clear image can be formed on the screen. The separation between the two positions is measured to be 20 cm. Is the unknown lens convex or concave? What is the focal length? Hint: The solution of a quadratic equation ax +bx+c=0 is -b±yb² – 4ac 2a
- 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!Two converging lenses are separated by 27.0 cm. The focal length of each lens is 13.0 cm. An object is placed 37.0 cm to the left of the lens that is on the left. Determine the final image distance relative to the lens on the right.