an object of size 5 cm is placed in front of a concave mirror at a dis

1.	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» 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 : $\implies {\rm{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}} \\ \\ \\ \implies \rm{\dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u}} \\ \\ \\ \implies \rm{\dfrac{1}{v} = \dfrac{-1}{10} - \bigg( \dfrac{-1}{20} \bigg) } \\ \\ \\ \implies \rm{\dfrac{1}{v} = \dfrac{-1}{10} + \dfrac{1}{20}} \\ \\ \\ \implies \rm{\dfrac{1}{v} = \dfrac{-2 + 1}{20}} \\ \\ \\ \implies \rm{\dfrac{1}{v} = \dfrac{-1}{20}} \\ \\ \\ \large {\implies{\boxed{ \sf{v \: = \: -20 \: cm}}}}$ ____________________ Now, use formula for magnification : $\implies \rm{m = \dfrac{-v}{u} = \dfrac{H'}{H_o}} \\ \\ \\ \implies \rm{\dfrac{-(-20)}{-20} = \dfrac{H'}{5}} \\ \\ \\ \implies \rm{\dfrac{20}{-20} = \dfrac{H'}{5}} \\ \\ \\ \implies \rm{-1 = \dfrac{H'}{5}} \\ \\ \\ \implies \rm{H' = 5 \times -1} \\ \\ \\ \large {\implies{\boxed{\sf{H' = -5 \: cm}}}}$ ____________________ $\bullet \: \: \: \: \rm{Nature\ of\ image\ is\ Real\ and\ inverted.}$ $\bullet \: \: \: \: \rm{Image\ is\ placed\ at\ distance\ of\ -20\ cm\ from\ mirror.}$ $\bullet \: \: \: \: \rm{Height\ of\ image\ is\ -5\ cm.}$

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»

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 :

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

____________________

Now, use formula for magnification :

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

____________________

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

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

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