A converging lens has a focal length of 87.0 cm. Locate the images for the follo

Question

A converging lens has a focal length of 87.0 cm. Locate the images for the following object distances, if they exist. Find the magnification. (Enter 0 in the q and M fields if no image exists.) (a) 87.0 cm q = cm M = Select all that apply to part (a). real virtual upright inverted no image (b) 19.3 cm q = cm M = Select all that apply to part (b). real virtual upright inverted no image (c) 435 cm q = cm M = Select all that apply to part (c). real virtual upright inverted no image

A converging lens has a focal length of 87.0 cm. Locate the images for the following object distances, if they exist. Find the magnification. (Enter 0 in the q and M fields if no image exists.) (a) 87.0 cm Cm Select all that apply to part (a). rea virtual upright inverted no image (b) 19.3 cm Cm Select all that apply to part (b). rea virtual upright inverted no image (c) 435 cm Cm Select all that apply to part (c). real virtual upright inverted no image

Explanation / Answer

(a) The object is at the focal point of the lens and therefore light from the object emerges from the lens as a parallel beam. There is therefore no image.
(b) The lens equation is
1/u + 1/v = 1/f
where u is the object distance, v is the image distance and f is the focal length. The convention I use is that distances to real objects and images are positive while distances to virtual objects and images are negative. The focal length of a convex lens is positive and the focal length of a concave lens is negative.
1/19.3 + 1/v = 1/87
1/v = 1/87 - 1/19.3 = -0.0403
v = -24.81 cm
The image is virtual and upright.
Magnification = v/u = 24.81/19.3 = 1.28
(c) 1/u + 1/v = 1/f
1/44 + 1/v = 1/11
1/v = 1/87 - 1/435 = 0.00919
v = 108.81 cm
The image is real and inverted.
magnification = v/u = 108.81/87 = 1.25

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