Tag: problems on mirror and magnification formula

Magnification $= +1$ signifies that the image formed in a plane mirror is of same size as the object. Positive sign in the value of magnification signifies that image formed by a plane mirror is erect.

Given that, $\displaystyle R=-24$cm
$\displaystyle f=-12cm$ and $\displaystyle m=1.5$
By the lens formula,
$\displaystyle \frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } $
$\displaystyle \frac { 1 }{ 1.5u } +\frac { 1 }{ u } =-\frac { 1 }{ 12 } $
$\displaystyle \frac { 2.5 }{ 1.5u } =-\frac { 1 }{ 12 } $
or, $\displaystyle u=-20cm$

Magnification by a mirror         $m = \dfrac{-v}{u}$

The image produced is virtual and erect and also diminished.

Since the object and the image move in opposite directions, the position of the object should be in between $f$ and $2f$.
So, $f < x < 2f$

Since the magnification produced is positive and less than 1,the mirror is a convex mirror.

General equation for a spherical mirror says that:
$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

The magnification is given as,

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

$2 = \dfrac{{ - v}}{u}$

$v =  - 2u$

Ignoring the sign and using mirror formula, we get

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

$\dfrac{1}{{2u}} + \dfrac{1}{u} = \dfrac{1}{f}$

$\dfrac{{1 + 2}}{{2u}} = \dfrac{1}{f}$

$u = \dfrac{{3f}}{2}$

Here, difference between object distance and image distance is also$u$.

The size of the image is equal to size of the object for plane mirror.

So, magnification is equal to $1$ .