2. A 2.0-cm-tall object is located 30 cm from a lens whose front and back surfac

Question

2. A 2.0-cm-tall object is located 30 cm from a lens whose front and back surfaces both have radii of curvature R1 = R2 = 10.0 cm, as show in the diagram below. The index of refraction of the glass in the lens has a value of 1.5.
a. Mathematically calculate the location of the final image. Is the lens a converging or a diverging lens? Briefly explain your answer. (Hint: First, you will need to find the focal length of the lens.) (20 points)
b. What is the magnification of the lens and the height of the image? (5 points)
c. Is the image real or virtual? Briefly explain your answer. (5 points)
d. Is the image upright or inverted? Briefly explain your answer. (3 points)
e. Is the image larger than the object or smaller than the object? Briefly explain your answer. (2 points)

Explanation / Answer

a) let, Focal length of lens be f, radius of curvature of front and back surface = R, length of object from aperture be u; length of image from aperture be v; height of object be h1; height of image be h2. now , 1/f = 1/R1 + 1/R2 =1/R + 1/R = 2/R =2/10 i.e f=R/2 = 5 cm. Focal length is positive implies the lens is a converging lens or convex lens,a double-convex lens in this case indeed. 1/f = 1/u + 1/v 1/v = 1/f - 1/u = 1/5 -1/30 =1/6 i.e v = 6 cm. b) Magnification M = h2/h1 = - v/u i.e M = -6/30 = -1/5 => h2 = M*h1 = -2/5 cm. negative value of image indicates that the image is inverted with respect to the object. c) a convex lens produces virtual image of an object only when the distance of object from aperture of lens is less than the focal length i.e object lies within the focal length of lens. In this case since the object is placed beyond the focal length we get a real image. d)image is inverted. e)Negative value of magnification obtained implies image is smaller than the original object.

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