Tag: mirror formula and magnification

$\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

Here, $m =\dfrac{v}{u} =-4$

$\Rightarrow u = \dfrac{-v}{4}$------------(1)

Also, $|u| + |v| = 1.5$

$\dfrac{v}{4} + v = 1.5$

$\Rightarrow \dfrac{(v+4 v)}{4} = 1.5$

 $\Rightarrow  v = 1.2 m$

So, putting the value of $v$ in equation (1):
 $u= \dfrac{-1.2}{4} = -0.3 m\,\,\,$

$\therefore  f =\dfrac{uv}{u-v}$

$\Rightarrow f=\dfrac{(-0.3 \times 1.2)}{(-0.3 - 1.2)} = 0.24m$
Hence the correct option is $(C)$

$\dfrac{2}{R}=\dfrac{1}{u}+\dfrac{1}{v}$