Question

At what distance from a concave mirror of focal length 20 cm should an object 2 cm long be placed in order to get an erect image 8 cm tall?​

Answer 1

Focal length of concave mirror (f) = -20 cm

$Height \: \: of \: \: the \: \: object \: (h _{o} ) \: = 2cm \\ Height \: \: of \: \: the \: \: image \: ( h_{i}) \: = 8cm \\ we \: \: need \: \: to \: \: find \: \: the \: \: object \: \: distance \: \: \\ so \: \: we \: \: known \: \: that \: \: \\ magnification \: \: m \: \: = \frac{h_{i}}{h_{o}} \\ \frac{h_{i}}{h_{o}} = \frac{ \cancel8 \: ^{4} }{ \cancel2} \\ \\ \frac{h_{i}}{h_{o}} = 4(erect \: image) \: \\ \\ m = - \frac{ v}{u} \\ 4 = - \frac{v}{u} \\ v = - 4u \\ let \: \: us \: \: now \: \: consider \: \: mirror \: \: formular \: \\ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \\ \frac{1}{ - 4u} + \frac{1}{u} + \frac{1}{ - 20} \\ u = \frac{ - 4}{20} \\ u = \frac{ - \cancel20 \: ^{5} }{ \cancel4} \: \\ u = - 5cm \\ \:$

So at -5cm distance from a concave mirror of focal length 20 cm should an object 2 cm long be placed in order to get an erect image 8 cm tall..

Answer 2

$\mathfrak \purple{Hii \: \: Army \: or \: blink \: }$

The object should be placed at a distance of 15

cm from the concave mirror.

Explanation :

We have :

→ Focal length (f) = 20 cm

Height of object (h) = 2 cm

Height of image (hi) = 8 cm

From the formula of magnification of a spherical mirror, we have :

$m = - ( \frac{v}{u} ) = \frac{h _{i}}{h _{o}} \:$

→ - (v / u) = 8/2

⇒ v / u = -4

⇒ V = -4u ---(1)

According to sign convention, we know that f and u are always negative for a concave mirror.

So, now putting values in the mirror formula, we

get :

1/v + 1/u = 1/f

1 / (- 4u) + 1 / u = 1 / (- 20)

→ -1 / 4u + 1 / u = -1/20

(-1 + 4) / 4u = -1/20

→ 3 / 4u = -1/20

⇒ -4u = 60

→ u = 60 / -4

→ u = -15 cm

$\mathfrak \purple{hope \: \: it \: \: helps \: \: u } \:$