An object is placed a distance of 3.40 f from a converging lens, where f is the
ID: 1836073 • Letter: A
Question
An object is placed a distance of 3.40f from a converging lens, where f is the lens's focal length. (Include the sign of the value in your answers.) (a) What is the location of the image formed by the lens?di = f
(b) Is the image real or virtual? realvirtual
(c) What is the magnification of the image?
(d) Is the image upright or inverted? uprightinverted An object is placed a distance of 3.40f from a converging lens, where f is the lens's focal length. (Include the sign of the value in your answers.) (a) What is the location of the image formed by the lens?
di = f
(b) Is the image real or virtual? realvirtual
(c) What is the magnification of the image?
(d) Is the image upright or inverted? uprightinverted (a) What is the location of the image formed by the lens?
di = f
(b) Is the image real or virtual? realvirtual
(c) What is the magnification of the image?
(d) Is the image upright or inverted? uprightinverted di = f realvirtual uprightinverted
Explanation / Answer
(A) 1/v - 1/u = 1/f => 1/v + 1/3.4f = 1/f => 1/v = (1/f) (1 - 1/3.4) => v = 3.4f/(3.4 - 1) = 1.42 f
(B) Image is real
(C) magnification, m = 1.42/3.4 = 0.42
(D) m > 0 => image is upright.