A virtual object is located to the right of a divergent mirror. The object\'s di

Question

A virtual object is located to the right of a divergent mirror. The object's distance from the mirror and its focal length are shown in the figure below. The divergent mirror in this problem is a part of a lens/mirror system so the object in this problem may be either real or virtual. The image is virtual, erect, and smaller. virtual, erect, and larger. virtual, inverted, and smaller. real, erect, and smaller real, inverted, and larger. virtual, inverted, and larger. real, erect, and larger. real, inverted, and smaller. What is the image distance? -5/7 f -19/26 f -7/11 f -19/31 f 14/5 f -17/28 f -17/28 f -17/11 f -9/5 f -16/7 f 17/10 f What is the magnification? 23 6

Explanation / Answer

1/u + 1/v = 1/f

f<0 for divergent lens

u = -14/19f because object is virtual

1/v = 19/14f - 1/f

1/v = 19 - 14 /14f

1/v = 5/14f

v = 14f/5

iamge is real erect and larger

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