Question

An object is placed at distance 20 cm in front of a concave mirror of radius of curvature 30 cm. Find the
position, nature and magnification of the image.



best answer should mark as brainlist

Answer 1

Given:-

Object distance (u) = -20 cm

radius of curvature = -30 cm

To find :-

Nature, magnification and position of the image.

Solution:-

Focal length of mirror is given by :-

$\boxed{\sf{f = \dfrac{r}{2}}}$

$f = \dfrac {-30}{2}$

$f = -15 cm$

Let the image distance be v.

Now, by applying mirror formula :-

$\boxed{\sf{\dfrac{1}{f}=\dfrac{1}{v}+ \dfrac{1}{u}}}$

$\dfrac{-1}{15}= \dfrac{1}{v}-\dfrac{1}{20}$

$\dfrac{-1}{15}=\dfrac{1}{v}-\dfrac{1}{20}$

$\dfrac{1}{v}= \dfrac{-1}{15}+\dfrac{1}{20}$

$\dfrac{1}{v}=\dfrac{-4+3}{60}$

$\dfrac{1}{v}=\dfrac{-1}{60}$

$v = -60 cm$

Magnification :-

$M=\dfrac{-v}{u}$

$M = \dfrac{-(-60)}{-20}$

$M = \dfrac{60}{-20}$

$M = \dfrac {-60}{20}$

$M = -3$

Nature:-

hence, image formed is real and inverted.