Question

an object 5 cm in length is placed at a distance of 30 cm in front of a convex mirror of radius of curvature 40 cm . find the position of the image , its nature and size.​

Answer 1

Given :

▪ Height of object = 5cm

▪ Distance of object = 30cm

▪ Radius of curvature = 40cm

▪ Type of mirror : convex

To Find :

↗ Position of image

↗ Size of image

↗ Nature of image

Formula :

✴ Mirror formula :

$\bigstar\:\underline{\boxed{\bf{\pink{\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}=\dfrac{2}{R}}}}}$

✴ Lateral manification :

$\bigstar\:\underline{\boxed{\bf{\blue{m=-\dfrac{v}{u}=\dfrac{h'}{h}}}}}$

• u denotes distance of object
• v denotes distance of image
• f denotes focal length
• R denotes radius of curvature
• h denotes height of object
• h' denotes height of image

Calculation :

⭕ Position of image :

$\leadsto\sf\:\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{2}{R}\\ \\ \leadsto\sf\:\dfrac{1}{(-30)}+\dfrac{1}{v}=\dfrac{2}{40}\\ \\ \leadsto\sf\:\dfrac{1}{v}=\dfrac{1}{20}+\dfrac{1}{30}\\ \\ \leadsto\sf\: \dfrac{1}{v}=\dfrac{3+2}{60}\\ \\ \leadsto\sf\:v=\dfrac{60}{5}\\ \\ \leadsto\underline{\boxed{\bf{\green{v=12cm}}}}\:\gray{\bigstar}$

⭕ Size of image :

$\implies\sf\:m=-\dfrac{v}{u}=\dfrac{h'}{h}\\ \\ \implies\sf\:-\dfrac{12}{(-30)}=\dfrac{h'}{5}\\ \\ \implies\sf\:\dfrac{12}{6}=h'\\ \\ \implies\underline{\boxed{\bf{\gray{h'=2cm}}}}\:\orange{\bigstar}$

⭕ Nature of image :

• Virtual
• Erect
• Small

Answer 2

$\large{\red{ \bf{ \underline {\underline{Answer}}}}} \\ \\ \purple{ \sf{ \mapsto Imge \: Position = 12 \: cm \: (behind \: mirror)}} \\ \\ \purple{ \sf{ \mapsto Image \:Height = 2 \: cm }} \\ \\ \purple{ \sf{ \mapsto Imge \: Nature = Errect, \: Virtual \: and \: Diminished}}\\ \\ \mathrm{\green{ \underline{Given}}} \\ \\ \sf{ \rightsquigarrow Distance \: (u) = - 30 \: cm} \\ \\ \sf{ \rightsquigarrow Radius \: (R) = 40 \: cm} \\ \\ \sf{ \rightsquigarrow Object\: (h_o) = 5 \: cm}\\ \\ \mathrm {\blue{ \underline{ To \: Find}}} \\ \\ \sf{ \rightsquigarrow v =\:?} \\ \\ \sf{ \rightsquigarrow h_i = \: ?}$

• According to given question

$\sf{\underline{We \: have \: to \: write }} \\ \\ \sf{ \implies f = \frac{R}{2} } \\ \\ \sf{ \implies f = \frac{ \cancel{40}}{ \cancel{2} }} \\ \\ \sf{ \implies f = 20 \: cm} \\ \\ \sf{\underline{We \: have \:to \: use \: a \: mirror \: formula}} \\ \\ \sf{ \implies \frac{1}{v} = \frac{1}{u} = \frac{1}{f} } \\ \\ \sf{ \implies \frac{1}{v} = \frac{1}{f} = \frac{1}{u} } \\ \\ \sf{ \implies \frac{(u - f)}{uf} } \\ \\ \sf{ \implies v = \frac{uf}{u - f} } \\ \\ \sf{ \implies \frac{( - 30 \times 20)}{( - 30 - 20)} } \\ \\ \purple{ \implies}\purple{ \underline{ \boxed{ \sf{ v=12 \: cm }}}} \\ \\ \sf{ \implies m = \frac{h_i}{h_o} } \\ \\ \sf{ \implies \frac{ - v}{u} } \\ \\ \sf{ \implies \frac{ \cancel{ -} \cancel{ 12}}{ \cancel{-} \cancel{ 30}} } \\ \\ \sf \purple{ \implies} \purple{ \underline{ \boxed{ \sf{h _i = 2 }}}}$

❍ Image position 12 cm (behind mirror)

❍ Image height : 2 cm

❍ Image nature : erect, virtual and diminished