How far from a lens of focal length 50.0 mm must the object be placed if it is to form a virtual image magnified in size by a factor of three?
A) 38.3 mm
B) 54.4 mm
C) 33.3 mm
D) 48.0 mm
E) 42.2 mm

Question

Quizplus · Accepted Answer

For a lens (whether converging or diverging), the magnification (m) is related to the object distance (u) and the image distance (v) by the formula $m = -\frac{v}{u}$. For a virtual image, magnification is positive and greater than 1 (in this case, 3). The lens formula also relates these distances to the focal length (f) by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$. Given that the magnification is 3, we have $m = 3 = -\frac{v}{u}$, implying $v = -3u$. Substituting $v = -3u$ into the lens formula with $f = 50.0$ mm gives $\frac{1}{50} = \frac{1}{-3u} + \frac{1}{u}$, which simplifies to find $u \approx 33.3$ mm.

How Far from a Lens of Focal Length 50