Question

An object 2.0cm high in placed 20.cm in front concave mirror of focal length 10.0cm .find the distance from the mirror at which screen should be placed in order to obtain a sharp image. What will be the size and the nature of the image formed?

Answer 1

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

• Object distance (u) = -15 cm

• Focal length (f) = -10 cm

• Height of the object (h) = 2 cm

• Height of the image (h') = ? cm

• Image distance (v) = ? cm

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

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

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

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

• Find the distance from the mirror at which screen should be placed in order to obtain a sharp image. What will be the size and the nature of the image formed?

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

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

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

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

• $\bold{\frac{1}{Image\:distance\:(\frac{1}{v} )} \:=\:\frac{1}{-10}\: -\: \frac{1}{-20}}$

• $\bold{\frac{1}{Image\:distance\:(\frac{1}{v} )} \:=\:\frac{1}{-10}\: +\: \frac{1}{20}}$  

• $\bold{\frac{1}{Image\:distance\:(\frac{1}{v} )} \:=\:\frac{-2+1}{20}}$  

• $\bold{\frac{1}{Image\:distance\:(\frac{1}{v} )} \:=\:\frac{-1}{20}}$  

• Image distance (v) = -20 cm

   

    ${\bold{{\underline{\red{To}\pink{\:find}\green{\:image\:height(h')}\purple{:}}}}}$

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

• Magnification (m) = $\bold{\frac{-\:(-20)}{-20}}$

• Magnification (m) = $\bold{\frac{1}{-1}}$

• Magnification (m) = -1 cm

Also,

• Magnification (m) = $\bold{\frac{height\:of \:image\:(h')}{height\:of\:object\:(h)} }$

• -1 = $\bold{\frac{height\:of \:image\:(h')}{2} }$

• $\bold{\frac{height\:of \:image\:(h')}{2} }$  = -1

• Height of image (h') = -1 x 2

• Height of image (h') = -2 cm

    ${\bold{{\underline{\red{Nature}\pink{\:of}\green{\:image}\purple{:}}}}}$

    The nature of image is:

• Real and inverted
• Same size of the object and image
• Object was placed at centre of curvature (C) and image was formed at centre of curvature (C)