Question

An object $\text{AB}$ of length $1\text{mm}$ is placed on the axis of concave mirror of focal length $10\text{cm}$. If end $\text{A}$ of the object is at $30\text{cm}$ from the mirror, then length of the image is

• A. $25\times {10}^{-5}\text{m}$
• B. $12.5\times {10}^{-5}\text{m}$
• C. $50\times {10}^{-4}\text{m}$
• D. $50\times {10}^{-3}\text{m}$

Answer 1

The correct option is A $25\times {10}^{-5}\text{m}$


As the length of the object is very less we can use the formula for longitudinal magnification:

$m_L=-{\left(m_{lateral}\right)}^2$

$m_L=-\frac{v^2}{u^2}$

Given:

$u=-30\text{cm};f=-10\text{cm}$

Using mirror formula,

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

$\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{-1}{10}-\left(\frac{1}{-30}\right)=\frac{-1}{10}+\frac{1}{30}$

$\Rightarrow v=-15\text{cm}$

$\therefore m_L=\frac{dv}{du}=-m_T^2=-{\left(\frac{-(-15)}{-30}\right)}^2=-\frac{1}{4}$

$\therefore \left|\frac{dv}{du}\right|=\frac{1}{4}$

$\Rightarrow \left|dv\right|=\frac{1}{4}\left|du\right|=\frac{1}{4}\times {10}^{-3}=25\times {10}^{-5}\text{m}$

Hence option (a) is correct choice.