# Problem: A candle (ho = 0.39 m) is placed to the left of a diverging lens (f = -0.054 m). The candle is so = 0.34 m to the left of the lens.ho = 0.39 mf = -0.054 mso = 0.34 mPart (a) Write an expression for the image distance, si.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, hi?

###### FREE Expert Solution

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 si:

si = (1/f - 1/so)-1 = ((so - f)/f•so)-1 = fso/(so - f)

92% (488 ratings) ###### Problem Details

A candle (ho = 0.39 m) is placed to the left of a diverging lens (f = -0.054 m). The candle is so = 0.34 m to the left of the lens.

ho = 0.39 m
f = -0.054 m
so = 0.34 m

Part (a) Write an expression for the image distance, si.

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, hi?