Lens equation:

$\overline{)\frac{\mathbf{1}}{\mathbf{f}}{\mathbf{=}}\frac{\mathbf{1}}{{\mathbf{s}}_{\mathbf{o}}}{\mathbf{+}}\frac{\mathbf{1}}{{\mathbf{s}}_{\mathbf{i}}}}$

Magnification:

$\overline{){\mathbf{m}}{\mathbf{=}}\frac{{\mathbf{h}}_{\mathbf{i}}}{{\mathbf{h}}_{\mathbf{o}}}{\mathbf{=}}{\mathbf{-}}\frac{{\mathbf{s}}_{\mathbf{i}}}{{\mathbf{s}}_{\mathbf{o}}}}$

**(a)**

Solve for s_{i}:

s_{i} = (1/f - 1/s_{o})^{-1} = ((s_{o} - f)/f•s_{o})^{-1} = fs_{o}/(s_{o} - f)

A candle (h_{o} = 0.39 m) is placed to the left of a diverging lens (f = -0.054 m). The candle is s_{o} = 0.34 m to the left of the lens.

*h*_{o} = 0.39 m*f* = -0.054 m*s*_{o} = 0.34 m

Part (a) Write an expression for the image distance, *s _{i}*.

Part (b) Numerically, what is the image distance in meters?

Part (c) Is this real or virtual?

Part (d) Numerically, what is the image height, *h _{i}*?