A diverging lens (f=-15.0 cm) is located 21.0 cm to the left of a converging lens (f=36.0 cm). A 3.60-cm-tall object stands to the left of the diverging lens, exactly at its focal point. (a) Determine the distance of the final image relative to the converging lens. (b) What is the height of the final image (including the proper algebraic sign)? (a) Number Units (b) Number i PUUnits Object Lens 1 Fi Lens 2
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- An object of height 10.0 cm is placed 20.0 cm away from a converging lens. The image is 30.0 cm high. One can conclude that the focal length of the lens can be (A) 12.0 cm only. (B) either 12.0 cm or 60.0 cm. (C) 15.0 cm only. (D) either 15.0 cm or 30.0 cm.= A leaf of length his positioned d below. Leaf d 70.3 cm in front of a converging lens with a focal length of 34.0 cm. An observer views the image of the leaf from a position 1.26 m behind the lens, as shown in the figure 1.26 m (a) What is the magnitude of the lateral magnification (the ratio of image size to the object size) produced by the lens? (b) What angular magnification is achieved by viewing the image of the leaf rather than viewing the leaf directly?In a two-lens system, the focl length of the lens is f1 = 4.00 cm, and that of then lens 2 is f2 = 8.00 cm. The distance between the lenses is 16 cm. If an object O is 2.00 cm high and is placed 6.00 cm in front of the lens 1, where is the final image through the two-lens system?
- an object of height 5 cm is placed 20 cm in front of a converging lens at focal length 10 cm. Behind the converging lens, and 25cm from it, there is a diverging lens of the focal length of 6 cm. Find the location of the final image, in centimeters, with respect to teh diverging lens. what is the magnification of the final image? what is the height of the final image?(a) A 1.00 cm-high object is placed 4.35 cm to the left of a converging lens of focal length 7.85 cm. A diverging lens of focal length -16.00 cm is 6.00 cm to the right of the converging lens. Find the position and height of the final image. cm---Select--- position height Is the image inverted or upright? upright inverted Is the image real or virtual? real virtual cm (b) What If? If an image of opposite characteristic, i.e., virtual if the image in part (a) is real and real if the image in part (a) is virtual, is to be obtained, what is the minimum distance (in cm), and in which direction, that the object must be moved from its original position? cm distance direction ---Select---An object of 1 cm tall is placed 3cm in front of a converging lens of focal length of 2 cm. (a) Use ray tracing to find the image. (b) Use the lens equation to find the image distance and compare it to your ray tracing and find the percentage error.
- 2,2A system of two lenses forms an image of an arrow at x = x3 = 57.4 cm. The first lens is a diverging lens located at x = 0 and has a focal length of magnitude f₁ = 11.5 cm. The second lens is located at x = x₂ = 25.4 cm and has an unknown focal length. The tip of the object arrow is located at (x,y) = (xo, Yo) = (-36 cm, 20.6 cm). (x,y) 1) What is x₁, the x-coordinate of image of the arrow formed by the first lens? cm Submit 2) What is y₁, the y-coordinate of the image of the tip of the arrow formed by the first lens? cm Submit X₂ Real and Inverted Real and Upright Virtual and Inverted Virtual and Upright cm Submit (+) 3) What is f2, the focal length of the second lens. If the lens is a converging lens, f2 is positive. If the lens is a diverging lens, f2 is negative. cm Submit + 4) What is y3, the y-coordinate of the image of the tip of the arrow formed by the two lens system? (+ 5) The positions of the two lenses are now interchnaged (i.e., the second lens is moved to x = 0 and the…(a) For a diverging lens (f = - 16.0 cm), find the image distance for an object that is 16.0 cm from the lens. (b) Determine the magnification of the lens.