Object and Image Distance from Image height, Object height, and focal length [closed]

Question

Is there a way to find the object and image distances from the object height, image height, and focal length? I understand that the magnification is equal to $-\frac{d_i}{d_o}$ or $\frac{h_i}{h_o}$, but how does that relate to the mirror equation, $\frac{1}{f} = \frac{1}{d_i} + \frac{1}{d_o}$?

Answer 1

To begin with,

\begin{align*} & f \text{ represents the Focal Length of the mirror.} \\ & v \text{ represents the Image Distance, which is the distance from the mirror to the image.} \\ & u \text{ represents the Object Distance, which is the distance from the mirror to the object.} \\ & h_i \text{ represents the Image Height.} \\ & h_o \text{ represents the Object Height.} \end{align*}

As per your question, we are provided with the values of the Object Height, Image Height, and the Focal Length of the mirror.

For simplicity's sake, suppose, \begin{align} \text{Image Height } (h_{i}) = 10cm \\ \text{Object Height } (h_{o}) = 5cm \\ \text{Focal Length } (f) = 10cm \end{align}

Then by the magnification formula,

\begin{align} &\text{Magnification } (M) = \frac{h_{i}}{h_{o}} = -\frac{v}{u} \\ &\implies \frac{h_{i}}{h_{o}} = -\frac{v}{u} \\ &\implies \frac{10cm}{5cm} = -\frac{v}{u} \\ &\implies 2 = -\frac{v}{u} \\ &\implies 2u = -v \\ &\implies v = -2u \quad \text{... (i)} \end{align}

Now by the mirror formula, \begin{align} &\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \\ &\implies \frac{1}{10} = \frac{1}{-2u} + \frac{1}{u} \quad \text{(v = -2u, from (i))} \\ &\implies \frac{1}{10} = \frac{1}{u} - \frac{1}{2u} \\ &\implies \frac{1}{10} = \frac{2 - 1}{2u} \\ &\implies \frac{1}{10} = \frac{1}{2u} \\ &\implies u = 5 \\ &∴ u = 5cm \\ &\implies v = -2u = -2 \times (5 cm) = -10 cm \end{align}

I hope you now understand how this works in a more general sense.