An object is held 85.3 cm away from a lens. The lens has a focal length of 43.2

Question

An object is held 85.3 cm away from a lens. The lens has a focal length of 43.2 cm.

a.) Determine the image distance.
cm

b.) Determine the magnification.


c.) If the object is 3.6 cm long, determine the image height.
cm

d.) Which of the following are true about the image? Choose all that apply.
The image is inverted.
The image is real.
The image is virtual.
The image is upright.


e.) Which of the following are true about the lens? Choose all that apply.
The lens is diverging.
The lens is concave.
The lens is converging.
The lens is convex.

Explanation / Answer

Object distance, do = 85.3 cm Focal length, f = 43.2 cm Object height, ho = 3.6 cm Let Image height = di Magnification = m (a) 1/f = (1 /do) + (1/di) 1/di= (1/f) - (1/do) = (1/43.2) - (1/85.3) = 0.023 - 0.0117 Image distance, di = 88.5 cm (b) Magnification, m = di / do = 88.5 / 85.3 = 1.04 (c) Height of the image, hi = m * ho = 1.04 * 3.6 = 3.73 cm (d) The image is upright because, the image height is positive. (e) Nature of the lens: Convex or converging [ since focal length is positive ]

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