Question

A concave mirror of focal length $\text{'}f\text{'}$ forms an erect image twice the size of the object. Find the object distance from the mirror.

Answer 1

Step 1: Given data

Given in the question, the focal length of the concave mirror = $f$.

Image size is twice the size of the object $(h_i=2\times h_0)$ , where $h_i$= size of the image and $h_o$= size of the object.

Magnification will be positive here as it forms an erect image.

Step 2: Formula to be used

The formula used

1. $\frac{h_i}{h_0}=m$ where $m$=magnification.
2. $m=\frac{-v}{u}$ where $u$= distance of the object from the mirror and $v$=distance of the image from the mirror.
3. Mirror formula is, $\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ where, $f$ is the focal length, $u$ is the object distance from the mirror and $v$ is the image distance from the mirror.

Step 3: Finding the object distance:

Now, solving the given condition,$(h_i=2\times h_0)$

$$\begin{gathered} h_i=2\times h_0 \\ \Rightarrow \frac{h_i}{h_0}=2=m=\frac{-v}{u} \\ \Rightarrow v=-2u \end{gathered}$$

Now putting the value in the mirror formula,

$$\begin{gathered} \frac{1}{f}=\frac{1}{u}+\frac{1}{-2u} \\ \Rightarrow \frac{1}{f}=\frac{2-1}{2u} \\ \Rightarrow \frac{1}{f}=\frac{1}{2u} \\ \therefore u=\frac{f}{2} \end{gathered}$$

Final answer: Therefore, the object's distance from the mirror is $\frac{f}{2}$.