Question

solve this question

no scam pls....​

Answer 1

$\bold{\blue{\underline{\red{G}\pink{iv}\green{en}\purple{:-}}}}$

Focal length (f) = 10 cm

Object distance (u) = - 5 cm

Let 'v' be the image distance and 'm' be the magnification.

$\bold{{\underline{\red{Formulas}\pink{\:to\:be}\green{\:used}\purple{:-}}}}$

• Magnification (m) = $\bold{\frac{-\: Image\:distance\:(-v)}{Object\:distance\:(u)}}$

• $\bold{\frac{1}{Image\:distance\:(\frac{1}{v} )}\: +\: \frac{1}{Object\:distance\:(\frac{1}{u} )} \:=\:\frac{1}{Focal\:length\:(\frac{1}{f} )} }$

${\bold{{\underline{\red{Q}\pink{uest}\green{ion}\purple{:-}}}}}$

• Find the position and magnification of the image and state its nature ?

${\bold{{\underline{\red{S}\pink{olut}\green{ion}\purple{:-}}}}}$

• $\bold{\frac{1}{Image\:distance\:(\frac{1}{v} )}\: +\: \frac{1}{Object\:distance\:(\frac{1}{u} )} \:=\:\frac{1}{Focal\:length\:(\frac{1}{f} )} }$

• $\bold{\frac{1}{Image\:distance\:(\frac{1}{v} )}\: +\: \frac{1}{-5} \:=\:\frac{1}{10} }$

• $\bold{\frac{1}{Image\:distance\:(\frac{1}{v} )}\:=\:\frac{1}{10} }\: -\: \frac{1}{-5}$

• $\bold{\frac{1}{Image\:distance\:(\frac{1}{v} )}\:=\:\frac{1}{10} }\: +\: \frac{1}{5}$

• $\bold{\frac{1}{Image\:distance\:(\frac{1}{v} )}\:=\:\frac{1\:+\:2}{10}}$

• $\bold{\frac{1}{Image\:distance\:(\frac{1}{v} )}\:=\:\frac{3}{10}}$

• Image distance (v) = 3.33 cm

• Magnification (m) = $\bold{\frac{-\: Image\:distance\:(-v)}{Object\:distance\:(u)}}$

• Magnification (m) = $\bold{\frac{-3.33}{-5}}$

• Magnification (m) = 0.666 cm

∴ The image is virtual and erect and formed 3.33 cm behind the mirror and diminished by 0.666 cm.