Question

A linear object of height $10 \, cm$ is kept in front of a concave mirror of radius of curvature $15 \, cm$ , at a distance of $10 \, cm$ . The image formed is

• A. Magnified and erect
• B. Magnified and inverted
• C. Diminished and erect
• D. Diminished and inverted

Answer 1

Answer: (B)

Given,
Height of object $\left(h_{0}\right)=10 \, cm$
Radius of curvature $\left(R\right)=15 \, cm$
Distance of object $\left(u\right)=-10 \, cm$
We know that,
$f=\frac{R}{2}=\frac{15}{2} \, cm$
According to mirror formula,
$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
$\frac{1}{- \frac{15}{2}}=\frac{1}{- 10}+\frac{1}{v}$
$\Rightarrow \, \, \frac{1}{v}=\frac{- 2}{15}+\frac{1}{10}$
$\Rightarrow \, \, \frac{1}{v}=\frac{- 4 + 3}{30}$
$\Rightarrow \, \, \frac{1}{v}=\frac{- 1}{30}$
$\Rightarrow \, \, v=-30 \, cm$
Magnification $\left(m\right)=\frac{h_{i}}{h_{o}}=\frac{- v}{u}$
$=\frac{h_{i}}{10}=\frac{- \left(- 30\right)}{- 10}$
$\Rightarrow \, \, h_{i}=-3$
Hence, image will be magnified and inverted.