Question

When an object is placed at a distance of $25\mathrm{cm}$ from a mirror, the magnification is $m_1$. The object is moved $15\mathrm{cm}$ away with respect to the earlier position, magnification becomes ${\mathrm{m}}_2$, If $\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}=4$, The focal length of the mirror is?

Answer 1

Step 1: Given data

$$\begin{gathered} {\mathrm{u}}_1=-25\mathrm{cm} \\ {\mathrm{u}}_2=-(25+15)\mathrm{cm} \\ =-40\mathrm{cm} \end{gathered}$$

$\frac{m_1}{m_2}=4$

Step 2: Formula used

Mirror Equation

1. The object distance ($u$), image distance ($v$), and focal length ($f$) are all expressed quantitatively in the mirror equation.
2. The magnification equation connects the image and object distance ratios to the image.

The following is the mirror equation:

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

Here, u is the object distance

v is the image distance

f is the focal length.

The following is the magnification equation:

$\mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}$

where m is the magnification.

Step 3: Calculating the focal length,

initial Magnification is:

${\mathrm{m}}_1=-\frac{{\mathrm{v}}_1}{{\mathrm{u}}_1}$

After modification magnification will be:

${\mathrm{m}}_2=-\frac{{\mathrm{v}}_2}{{\mathrm{u}}_2}$

As given$\frac{m_1}{m_2}=4$, then

$\frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}=\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}\times \frac{{\mathrm{u}}_2}{{\mathrm{u}}_1}$

$4=\frac{{\mathrm{v}}_1}{{\mathrm{v}}_2}\times \frac{40}{25}$

${\mathrm{v}}_1=2.5{\mathrm{v}}_2$

By putting the value of $u_{1}and{v}_1$in the mirror equation:

$$\begin{gathered} \Rightarrow \frac{1}{{\mathrm{u}}_1}+\frac{1}{{\mathrm{v}}_1}=\frac{1}{\mathrm{f}} \\ \frac{1}{-25}+\frac{1}{2.5{\mathrm{v}}_2}=\frac{1}{\mathrm{f}}...............\left(1\right) \end{gathered}$$

By putting the value of $u_{2}and{v}_2$ in the mirror equation:

$$\begin{gathered} \Rightarrow \frac{1}{{\mathrm{u}}_2}+\frac{1}{{\mathrm{v}}_2}=\frac{1}{\mathrm{f}} \\ \frac{1}{-40}+\frac{1}{{\mathrm{v}}_2}=\frac{1}{\mathrm{f}}...................\left(2\right) \end{gathered}$$

By dividing equation ($2$) by $2.5$,

$\frac{1}{-100}+\frac{1}{2.5v_2}=\frac{1}{f}......................................\left(3\right)$

By subtracting equation ($3$) from ($1$)

$\frac{1}{-25}+\frac{1}{100}=\frac{1}{\mathrm{f}}-\frac{1}{2.5\mathrm{f}}$

$$\begin{gathered} \frac{3}{-100}=\frac{1.5}{2.5\mathrm{f}} \\ \mathrm{f}=-20\mathrm{cm} \end{gathered}$$

Therefore, the focal length of the mirror is $-20\mathrm{cm}$. The negative sign implies that it is a concave mirror.