Question

A thin stick of length $\frac{f}{5}$ is placed along the principal axis of a concave mirror of focal length f such that its image which is real and elongated just touches the stick. what is the magnification?

Answer 1

Step 1, Given

Length of the stick = $\frac{f}{5}$

Focal length = f

The figure of the given situation

Step 2,

From the mirror formula

$\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

Or, $\frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

Or, $\frac{1}{v}=\frac{1}{f}-\frac{1}{\frac{5f}{3}}$

After solving

$v=\frac{5f}{2}$

Now from the figure, we can write

$A^{\prime }{B}^{\prime }=\frac{5f}{2}-2f=\frac{f}{2}$

And $AB=\frac{f}{3}$

Now we know

Magnification $M=\frac{Sizeoftheimage}{Sizeoftheobject}=\frac{\frac{f}{2}}{\frac{f}{3}}=\frac{3}{2}$

Hence the magnification is equal to 1.5.