Physics - Geometrical Optics Question with Solution | TestHub

According to new cartesian sign convention, Object distance, $u=-15\mathrm{cm}$ Focal length of the concave mirror, $f=-10\mathrm{cm}$ Height of the object, $h_O=2\mathrm{cm}$ According to mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ $\Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-10}-\frac{1}{-15}=\frac{1}{-10}+\frac{1}{15}$ or $v=-30\mathrm{cm}$. The image is formed at a distance of $30\mathrm{cm}$ from the mirror on the same side of the object. It is a real image. Magnification of the mirror, $m=\frac{-v}{u}=\frac{h_I}{h_O}$ $\Rightarrow \frac{-(-30)}{-15}=\frac{h_I}{2}\Rightarrow h_I=-4\mathrm{cm}$. Negative sign shows that image is inverted. The image is real, inverted, of size $4\mathrm{cm}$ at a distance of $30\mathrm{cm}$ in front of the mirror.