Answer to Question #126475 in Mechanics | Relativity for Stefanie Hernandez-Mendez

Question

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?

Answer 1

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