Question

A linear object of height $10cm$ is kept in front of a concave mirror of radius of curvature $15cm$ , at a distance of $10cm$ . 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 $\left(h_0\right)=10cm$
Radius of curvature $\left(R\right)=15cm$
Distance of object $\left(u\right)=-10cm$
We know that,
$f=\frac{R}{2}=\frac{15}{2}cm$
According to mirror formula,
$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$
$\frac{1}{-\frac{15}{2}}=\frac{1}{-10}+\frac{1}{v}$
$\Rightarrow \frac{1}{v}=\frac{-2}{15}+\frac{1}{10}$
$\Rightarrow \frac{1}{v}=\frac{-4+3}{30}$
$\Rightarrow \frac{1}{v}=\frac{-1}{30}$
$\Rightarrow v=-30cm$
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.