Tag: physics

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} $