Question

A concave mirror producing magnification of 3cm when object is placed at 15cm. Find focal length of the concave mirror

Answer 1

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Answer 2

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• Magnification (m) = 3 cm

• Object distance (u) = - 15 cm

• Let 'v' be the image distance and 'f' be the focal length .

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

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

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• To find the focal length of concave mirror ?

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

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

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

• 3 = $\bold{\frac{-\: Image\:distance\:(-v)}{-15}}$

• 3 × (-15) = - Image distance (-v)  
• - Image distance (-v) = -45
• ∴ Image distance (v) = 45 cm

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

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

• $\bold{\frac{1}{45}\: +\: \frac{1}{15} \:=\:\frac{1}{Focal\:length\:(\frac{1}{f} )} }$  

• $\bold{\frac{3\:+\:1}{45}\:=\:\frac{1}{Focal\:length\:(\frac{1}{f} )} }$

• $\bold{\frac{1}{Focal\:length\:(\frac{1}{f} )}\:=\:\frac{4}{45}}$

• ∴ Focal length (f) = 11.25 cm