Question

an object of height 5 cm is placed at 30cm distance of the principal axis front of a concave mirror of focal length 20cm find the image distance and size of the image?​

Answer 1

Here,

• Object distance, u = - 30 cm
• Focal length, f = - 20 cm
• Height of object, $\sf h_o$ = 5 cm

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We have to find,

• Image distance, v = ?
• Size/Height of image, $\bf h_i$ = ?

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${\underline{\sf{\bigstar\;Using\;mirror\;formula\;:}}}$

$\star\;{\boxed{\sf{\purple{ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}}}$

$\qquad\: \: \: :\implies\sf \dfrac{1}{-20} = \dfrac{1}{-30} + \dfrac{1}{v}$

$\qquad:\implies\sf \dfrac{1}{v} = \dfrac{1}{-20} - \bigg( \dfrac{1}{-30} \bigg)$

$\qquad\qquad:\implies\sf \dfrac{1}{v} = \dfrac{1}{-20} + \dfrac{1}{30}$

$\qquad\qquad\quad:\implies\sf \dfrac{1}{v} = \dfrac{- 3 + 2}{60}$

$\qquad\qquad\qquad \: \: :\implies\sf \dfrac{1}{v} = \dfrac{- 1}{60}$

$\qquad\qquad:\implies{\boxed{\frak{\pink{v = - 60\;cm}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;image\;distance\;is\;60\;cm\;from\;mirror.}}}$

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${\underline{\sf{\bigstar\;Now,\;using\; magnification\;formula\;to\;find\;size\;of\;image,}}}$

$\star\;{\boxed{\sf{\purple{m = \dfrac{- v}{u}}}}}$

$\qquad \: :\implies\sf m = \dfrac{- (-60)}{-30}$

$\qquad \: \: \: \: \: \: :\implies\sf m = \cancel{ \dfrac{60}{-30}}$

$\qquad \: :\implies{\boxed{\frak{\pink{m = - 2}}}}\;\bigstar$

Again,

$\star\;{\boxed{\sf{\purple{m = \dfrac{h_i}{h_o}}}}}$

$\qquad\quad:\implies\sf - 2 = \dfrac{h_i}{5}$

$\qquad:\implies\sf - 2 \times 5 = h_i$

$\qquad:\implies{\boxed{\frak{\pink{h_i = - 10}}}}\;\bigstar$

$\therefore\;{\underline{\sf{Hence,\;the\;size\;of\;image\;is\; \bf{-10\;cm}.}}}$

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${\underline{\sf{\bigstar\;Nature\;of\;image\;:}}}$

• Real

• Inverted with a height of 10 cm.