Question

At what distance from a concave mirror of focal length 10cm and an object 2cm higher is placed inoder to get an irrect image of 6cm tall

Answer 1

$\mathfrak{Hello \: there }$
$\bf{Given \: that }$

• Focal length , f = - 10 cm

• Height of the object , H = 2 cm

• Image height , h = + 6 cm

• Object distance , u =

$\bf \underline {using \: magnification \: formula } \\ \\ \rightarrow \boxed { \bf m = \frac{ h}{H} = \frac{ - v}{u} } \\ \\ \rightarrow \: \frac{6}{2} = \frac{ - v}{u} \rightarrow \boxed {v = - 3u}$

$\bf \underline{ \: using \: mirror \: formula} \\ \\ \rightarrow \boxed { \bf \: \frac{1}{f} = \frac{1}{v} + \frac{1}{u} } \\ \\ \rightarrow \: \frac{1}{u} = \frac{1}{ - 10} - \frac{1}{ - 3u} \rightarrow \frac{ - 1}{10} + \frac{1}{3u } \\ \\ \rightarrow \: \frac{1}{u} = \frac{ - 3u + 10}{30u} \\ \\ \rightarrow - 3u + 10 = 30 = > u = \frac{ - 20}{3} \\ \\ \rightarrow \boxed {\bf \:u = - 6.67}$

$\mathfrak{Hope \: this \: helps \: you }$

Answer 2

HEY!!

_____________________________

▶f = -10

▶h' = +6 

▶h = +2

▶h'/h = -v/u

▶6/2 = -v/u

▶-v/u = 3

▶v = -3u          ------------1

▶f = vu/v+u

▶-10 = (-3u)u/-3u+u

▶-10 = -3u^2/-2u

▶-3u = 20

▶u = -20/3

▶u = -6.67

✔Therefore, the object should be placed at 6.67 cm in front of the mirror.