Question

object distance 30cm image distance 25cm the image properties is​

Answer 1

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

• Object distance (u) = 30 cm
• Image distance (v) = 25 cm
• Let 'f' be the focal length 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 Properties of image formed ?

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

   ${\bold{{\underline{\red{To}\pink{\:find}\green{\:focal\:length}\purple{:}}}}}$

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

• $\bold{\frac{1}{25}\: +\: \frac{1}{-30} \:=\:\frac{1}{Focal\:length\:(\frac{1}{f} )} }$

• $\bold{\frac{1}{Focal\:length\:(\frac{1}{f} )} \:=\:\frac{1}{25}\: +\: \frac{1}{-30} }$

• $\bold{\frac{1}{Focal\:length\:(\frac{1}{f})}\:=\:\frac{-6\:+\: 5}{-150}}$  

• $\bold{\frac{1}{Focal\:length\:(\frac{1}{f})}\:=\:\frac{-1}{-150}}$

• $\bold{\frac{1}{Focal\:length\:(\frac{1}{f})}\:=\:\frac{1}{150}}$

• ∴ Focal length (f) = 150 cm

    ${\bold{{\underline{\red{To}\pink{\:find}\green{\:magnification}\purple{:}}}}}$

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

• Magnification (m) = $\bold{\frac{-25}{30} }$

• ∴ Magnification (m) = -0.833 cm

${\bold{{\underline{\red{O}\pink{bser}\green{vation}\purple{:-}}}}}$

∵ Object is placed between the Focus (F) and Pole (P) the image formed will be Virtual and ertect . Also, the image will be reduced in size by 0.833 cm .