Physics, asked by Mahitha234, 8 months ago

Plz can anyone answer my question? ​

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Answered by Anonymous
1

Here,

Object distance, u = - 30 cm

Focal length, f = - 20 cm

Height of object, \sf h_o = 5 cm

We have to find,

Image distance, v = ?

Size/Height of image, \bf h_i = ?

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━

{\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.

Answered by shrinkuashihuny
0

Answer:

answer is 60

Explanation:

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