An object is placed just slightly less than from a concave lens, and now you mov

Question

An object is placed just slightly less than from a concave lens, and now you move the object toward the lens. What happens to the virtual image? The image grows larger, and moves closer to the lens the image remains the same size, and moves farther from the lens the image grows smaller, and remains fixed the image grows smaller, and moves farther from the lens the image grows smaller, and moves closer to the lens the image remains the same size, and remains fixed the image grows larger, and remains fixed the image grows larger, and moves farther from the lens the image remains the same size, and moves closer to the lens In procedure 2, suppose the convex lens is placed at the 40.3 cm mark, and an image forms on the screen at the 58.8 cm mark. Find: the focal length of this lens. Cm the absolute value of the magnification of this lens the height of the image cm

Explanation / Answer

1) the image grows larger and moves closer to the lens

2) 1/f +1/40.3 = 1/18.5 => f = 34.2 cm

magnification = -18.5/40.3 = -0.459

hi = 0.459 x (height of object)

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