Question

an object 5.0cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. find nature, position and size of image​

Answer 1

Answer:

u = -20 cm, h= 5 cm. Image is virtual and erect and formed behind the mirror.

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Answer 2

$\footnotesize\tt{Focal \: length = \frac{Radius \: of \: Curvature}{2}}$

$\footnotesize\tt{Focal \: length = \frac{30}{2}}$

$\footnotesize\tt{Focal \: length = \cancel\frac{30}{2}}$

$\footnotesize\tt{Focal \: length = 15cm}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\bullet\footnotesize\tt \purple{ \: Given:-}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{u = -20cm}$

$\footnotesize\tt{v = ?}$

$\footnotesize\tt{h1 = + 5 cm}$

$\footnotesize\tt{Convex \: lens}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\bullet\footnotesize\tt \purple{ \: Solution:-}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\tt{ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }$

$\tt{ \frac{1}{15} = \frac{1}{v} + \frac{1}{ - 20} }$

$\tt{ \frac{1}{15} = \frac{1}{v} - \frac{1}{ 20} }$

$\tt{ \frac{1}{15} + \frac{1}{ 20} = \frac{1}{v} }$

$\tt{ \frac{4 + 3}{60}= \frac{1}{v} }$

$\tt{ \frac{7}{60}= \frac{1}{v} }$

$\footnotesize \boxed{ \bullet\tt \purple{ \: v = \frac{60}{7} }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\bullet\footnotesize\tt \purple{ \: Behind \: the \: mirror}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{ Magnification (m)} \tt{ \: = \frac{ - v}{u} }$

$\footnotesize\tt{ Magnification (m)} \tt{ \: = \frac{ \frac{ - 60}{7} }{ - 20} }$

$\footnotesize\tt{ Magnification (m)} \tt{ \: = \frac{ - 60 }{ 7 \times - 20} }$

$\boxed{\bullet\footnotesize\tt \purple{ \: Magnification (m)} \tt \purple{ \: = \frac{ 3 }{ 7} }}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\bullet\footnotesize\tt \purple{ \: Virtual \: and \: Erect}$

$\bullet\footnotesize\tt \purple{ \: Diminished}$

$\: \: \: \: \: \: \: \: \: \: \: \: \:$

$\footnotesize\tt{ Magnification (m)} \tt{ \: = \frac{ h2}{h1} }$

$\tt{ \frac{ h2}{5} = \frac{ \frac{ - 60}{7} }{ - 20} }$

$\tt{ \frac{ h2}{5} = \frac{60}{7 \times 20} }$

$\footnotesize\boxed{ \bullet \: \footnotesize \tt \purple{ h2} \purple \tt \purple{ \: = \frac{15}{7} }}$