A lighted object is placed 45.5cm in front of a converging lens with a focal len

Question

A lighted object is placed 45.5cm in front of a converging lens with a focal length of 32.5 cm. The image of the object is focused on a screen and its size and orientation recorded. The lens is removed and a mirror is placed 65.5 cm from the object on the opposite side from the screen. The image from the mirror is also in focus without moving the screen. (Note: The figure may not be to scale, and the mirror and lend are used one at a time, not at the same time. They are both included to give you an idea of the setup.) a) Determine the position of the screen in reference to the lens. b) Find the magnification of the image formed by the lens. c) Determine the focal length of the mirror. d) Find the magnification of the image formed by the mirror. Which image (lens or mirror) was larger?

Explanation / Answer

a) 1/v - 1/u = 1/f => 1/v + 1/45.5 = 1/32.5 => 1/v = (45.5 - 32.5)/(45.5 x 32.5) => v = 113.75 cm (distance of screen from lens)

b) magnification, m = v/u = -2.50

c) Distance of image (on screen) from mirror, v = 65.5 + 45.5 + 113.75 = 224.75
1/v + 1/u = 1/f => 1/224.75 + 1/65.5 = 1/f = > f = 50.72 cm

d) magnification, m = - v/u = - 3.43

The image formed by mirror is larger.

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