As we know,

Magnifaction of a mirror is defined as
$m=\dfrac{size of image}{size of object}$
and since, in case of virtual and errect image, size of image and object both are positive  . Hence magnification created by mirror is positive.

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

Magnification of image created by mirror is defined as
$m=\dfrac{size of image}{size of object}$
and in case of inverted image. Size of image is negative whereas size of object is positive. Hence , magnification produced is negative and it can be unity when object is placed at center of curvature and infinity when object is at focus.

We know,
$ 1/f = 1/u +1/v$
Or $2/r = 1/u +1/v$
Since radius of curvature is double the focal length. This leads to option A.

$We\quad know\quad \dfrac { 1 }{ f } =\dfrac { 1 }{ u } +\dfrac { 1 }{ v } ;\quad \ where\quad f=focal\quad length=?\ u=\quad initial\quad distance=20\quad cm\ v=\quad final\quad distance=10cm\ \dfrac { 1 }{ f } =\dfrac { 1 }{ -20 } +\dfrac { 1 }{ (-10) } \ f=\dfrac { 20 }{ 3 } \quad cm\ Now\quad object\quad moved\quad towards\quad the\quad mirror=0.1\quad cm,so\quad the\quad new\quad value\quad of\quad u=9.9\quad cm\ Again\quad \quad \quad \ \dfrac { 1 }{ f } =\dfrac { 1 }{ { u } _{ 1 } } +\dfrac { 1 }{ { v } _{ 1 } } \ \dfrac { 1 }{ f } -\dfrac { 1 }{ { u } _{ 1 } } =\dfrac { 1 }{ { v } _{ 1 } } \ \dfrac { 1 }{ { v } _{ 1 } } =\quad \dfrac { 3 }{ 20 } -\dfrac { 1 }{ (-9.9) } \ { v } _{ 1 }=20.4\quad cm\ The\quad change\quad in\quad final\quad distance\quad =20.4cm\quad -20\quad cm=0.4\quad cm\ \ v=\dfrac { 9\times 1600 }{ 3600 } =\quad 4cm/s$

The magnification produced by a plane mirror is $+1$ implies that the image formed by a plane mirror is virtual$,$

The number of images formed by two mirrors at an angle $\theta $ is

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

Given: For a convex mirror, $R = -40 cm$.
$v = -10 cm$ (image is virtual).
From mirror formula, we have

$\displaystyle \frac {2}{R}=\frac {1}{u}+\frac {1}{v}$

$\displaystyle \frac {1}{u}=\frac {2}{R}-\frac {1}
{v}=\frac {2v-R}{vR}$

$\displaystyle u=\frac {vR}{2v-R}=\frac {(-10cm) \times (-40cm)}{2 \times (-10cm) - (-40cm)}$

$=+20cm$
Thus, object is placed $20 cm$ in front of the mirror.

Focal length, $f = \dfrac {R}{2}$, where $R$ = radius of curvature

The values of known parameter should be used with their proper sign convention. No sign should be attached to the unknown parameter during calculation. Its sign will come of its own after calculation.

According to mirror equation.

from mirror formula