Question #126475

A candle (ho = 0.22 m) is placed to the left of a diverging lens (f = -0.067 m). The candle is do = 0.36 m to the left of the lens.

Numerically, what is the image distance, di in meters?

Is this real or virtual?

Numerically, what is the image height, hi?

Numerically, what is the image distance, di in meters?

Is this real or virtual?

Numerically, what is the image height, hi?

Expert's answer

(1) The image is virtual.

(2) "\\frac{1}{d_0}-\\frac{1}{d_i}=-\\frac{1}{f}\\to \\frac{1}{0.36}-\\frac{1}{d_i}=-\\frac{1}{0.067}\\to d_i=0.056m"

(3) "\\frac{d_i}{d_0}=\\frac{h_i}{h_0}\\to h_i=\\frac{d_i}{d_0}\\cdot h_0=\\frac{0.056}{0.36}\\cdot0.22=0.034m"

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