An object of height $3cm$ is placed at a distance of $15cm$ infront of a concave mirror of radius of curvature $10cm$. Calculate the height of image and distance of image?

Given that,

Height $h=3cm$

Distance $u=-15cm$

Radius = $10cm$

Now, the focus

$f=\frac{r}{2}$

$f=\frac{10}{2}=5cm$

Now, by using mirror equation

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

$\frac{1}{5}=\frac{1}{v}-\frac{1}{15}$

$\frac{1}{v}=\frac{1}{5}+\frac{1}{15}$

$v=\frac{15}{4}cm$

$v=4cm$

So, the image will be formed between focus and pole

Now, the magnification is

$m=\frac{-v}{u}$

$m=\frac{4}{15}$

Now,

$m=\frac{h^{\prime }}{h}$

$\frac{4}{15}=\frac{h^{\prime }}{3}$

$h^{\prime }=\frac{12}{15}$

$h^{\prime }=0.8$

Hence, the height of image is $0.8cm$ and distance of image is $4cm$