Question

An object of height 12 cm is placed at a distance of 9 cm in front of a concave mirror of focal length 4 cm. Find the position and size of the image​

Answer 1

$\Large{\underline{\underline{\bf{Solution :}}}}$

★ Given :

Height of object (H) = 12 cm

Object Distance (u) = -9 cm

Focal Length (f) = -4 cm

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★ To Find :

We have to find the position and size of the image.

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★ Solution :

We know the mirror formula

$\Large{\implies{\boxed{\boxed{\sf{\frac{1}{f}= \frac{1}{v} + \frac{1}{u}}}}}}$

_____________[Put Values]

$\sf{→ \frac{-1}{4} = \frac{1}{v} + (\frac{-1}{9})} \\ \\ \sf{→ \frac{1}{v} = \frac{-1}{4} + \frac{1}{9}} \\ \\ \bf{Taking \: LCM} \\ \\ \sf{→ \frac{1}{v} = \frac{-9 + 4}{36}} \\ \\ \sf{→ \frac{1}{v} = \frac{-5}{36}} \\ \\ \sf{→ v = \frac{-36}{5}} \\ \\ \bf{→ v = -7.2} \\ \\ \Large{\implies{\boxed{\boxed{\sf{v = -7.2 \: cm}}}}}$

$\rule{200}{2}$

$\Large{\implies{\boxed{\boxed{\sf{\frac{-v}{u} = \frac{H'}{H}}}}}} \\ \\ \sf{→ \frac{7.2}{-9} = \frac{H'}{12}} \\ \\ \sf{→ H' = \frac{7.2 \times \cancel{12}}{-\cancel{9}}} \\ \\ \sf{→ H' = \frac{7.2 \times 4}{3}} \\ \\ \sf{→ H' = \frac{-28.8}{3}} \\ \\ \sf{→ H' = -9.6 } \\ \\ \Large{\implies{\boxed{\boxed{\sf{H' = -9.6 \: cm}}}}}$