Question

an object of size 5 cm is placed in front of a concave mirror at a distance of 20cm whose focal length is 10 centimetre find the position nature and size of image​

Answer 1

GiveN :

• Object Height $\sf{(H_o)\ =\ 5\ cm}$
• Object Distance $\sf{(u)\ =\ -20\ cm}$
• Focal length $\sf{(f)\ =\ -10\ cm}$

To FinD :

• Nature, Position, Size of image

SolutioN :

Use Mirror Formula :

$\implies {\rm{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}} \\ \\ \\ \implies \rm{\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}} \\ \\ \\ \implies \rm{\dfrac{1}{v} = \dfrac{-1}{10} - \bigg( \dfrac{-1}{20} \bigg) } \\ \\ \\ \implies \rm{\dfrac{1}{v} = \dfrac{-1}{10} + \dfrac{1}{20}} \\ \\ \\ \implies \rm{\dfrac{1}{v} = \dfrac{-2 + 1}{20}} \\ \\ \\ \implies \rm{\dfrac{1}{v} = \dfrac{-1}{20}} \\ \\ \\ \large {\implies{\boxed{ \sf{v \: = \: -20 \: cm}}}}$

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Now, use formula for magnification :

$\implies \rm{m = \dfrac{-v}{u} = \dfrac{H'}{H_o}} \\ \\ \\ \implies \rm{\dfrac{-(-20)}{-20} = \dfrac{H'}{5}} \\ \\ \\ \implies \rm{\dfrac{20}{-20} = \dfrac{H'}{5}} \\ \\ \\ \implies \rm{-1 = \dfrac{H'}{5}} \\ \\ \\ \implies \rm{H' = 5 \times -1} \\ \\ \\ \large {\implies{\boxed{\sf{H' = -5 \: cm}}}}$

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$\bullet \: \: \: \: \rm{Nature\ of\ image\ is\ Real\ and\ inverted.}$

$\bullet \: \: \: \: \rm{Image\ is\ placed\ at\ distance\ of\ -20\ cm\ from\ mirror.}$

$\bullet \: \: \: \: \rm{Height\ of\ image\ is\ -5\ cm.}$

Answer 2

In this question We used two Formula First one is mirror formula

1/f = 1/v + 1/u

Here f is focal length , v is image distance and u is object distance.

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Second one is Formula for magnification of image

H'/Ho = -v/u

Here H' is height of image , Ho is height of object .

Thanks :)