(B) Repeat the problem, when the object is placed at 10.0 cm. Locate the image b
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(B) Repeat the problem, when the object is placed at 10.0 cm. Locate the image by substituting into the thin-lens equation: 1/10.0 cm + 1/q = 1/10.0 cm rightarrow 1/q = 0 This equation is satisfied only in the limit as q becomes infinite. q rightarrow infinity (c) Repeat the problem when the object is placed 5.00 cm from the lens. See the ray diagram shown in figure b. Substitute into the thin-lens equation to locate the image. 1/5.00 cm + 1/q = 1/10.0 cm Solve for q, which is negative, meaning the image is on the same side as the object and is virtual. q = -10.0 cm Substitute the values of p and q into the magnification equation. M is positive and larger than 1, so the image is upright and double the object size. M = q/p = - (-10.0 cm/5.00 cm) = +2.00 LEARN MORE REMARKS The ability of a lens to magnify objects led to the inventions of reading glasses, microscopes, and telescopes. QUESTION If the lens is used to form an image of the sun on a screen, the lens should be located at a distance from the screen equal to: the distance from the lens to its focal point. a value between the distance to the focal point and twice the distance to the focal point. the radius of the surface of the lens. twice the distance from the lens to its focal point. the sum of the radii of the two lens surfaces. PRACTICE IT Use the worked example above to help you solve this problem. A converging lens of focal length 8.6 cm forms images of an object situated at various distances. (a) If the object is placed 25.8 cm from the lens, locate the image, state whether it's real or virtual, and find its magnification. (If either of the quantities evaluate to infinity, type INFINITY.) q = ______cm M = ____ (b) Repeat the problem when the object is at 8.6 cm. q = _______ cm M = _________ (c) Repeat again when the object is 4.30 cm from the lens. Q = ______ cm M = ______ Suppose the image of an object is upright and magnified 1.73 times when the object is placed 12.8 cm from a lens. Find the location of the image and the focal length of the lens. Q = ______ cm f = ________ cmExplanation / Answer
(1) Since the object is at infinity therefore the image will be formed at the focus, therefore the screen should be at focus.
hence the option will be a i.e the distance from the lens to its focal point.
(2)
object distance (do) = 25.8 cm
focal length (f) = 8.6 cm (since converging lens therefore positive)
We know image (di) = dof / (do-f) = 25.8*8.6 / (25.8 - 8.6) = 12.9 cm
Therefore image will be formed at 12.9 cm .
magnification = -di/do = -12.9/25.8 = -1/2
Hence the image would be real , dimnished and inverted.
(b) when object is at 8.6 cm that is at focus.
When object is at focus then we get image at infinity
di = dof / (do-f) = 8.6*8.6 / (8.6 - 8.6) = 8.62/0 = infinite
Magnification = -di/do = infinite
hence image will be infinitely large , real and inverted.
(c) do = 4.3 cm
di = 4.3*8.6 / (4.3-8.6) = -8.6 cm
magnification = -di/do = -(-8.6) /4.3 = 2
Hence image will be virtual , erect and magnfied.
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