Physics - Geometrical Optics Question with Solution | TestHub

Given,

Height of object$\left(h_0\right)=10\mathrm{ c}\mathrm{m}$

Radius of curvature$\left(R\right)=15\mathrm{ c}\mathrm{m}$

Distance of object$\left(u\right)=-10\mathrm{ c}\mathrm{m}$

We know that,

$f=\frac{R}{2}=\frac{15}{2}\mathrm{ c}\mathrm{m}$

According to mirror formula,

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

$\frac{1}{-\frac{15}{2}}=\frac{1}{-10}+\frac{1}{v}$

$\Rightarrow \frac{1}{v}=\frac{-2}{15}+\frac{1}{10}$

$\Rightarrow \frac{1}{v}=\frac{-4+3}{30}$

$\Rightarrow \frac{1}{v}=\frac{-1}{30}$

$\Rightarrow v=-30\mathrm{ c}\mathrm{m}$

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.