Question

An object of
5.0 cm in size is placed at a distance of 20.0 cm from a concave mirror of focal length of
15.0 cm . At what distance from the mirror should the screen be placed to obtain a sharp image? Also
calculate the size of the image.​

Answer 1

_________ [ Given ]

• h' = ( height of object ) = 5 cm

• u = ( object distance ) = - 20.0 cm = - 20 cm

• f = ( focal length ) = - 15 cm

_________ [ Find ]

• We have to find the distance of the image from mirror (v) and height of the image (h).

_________ [ Note ]

• Converging mirror means cancave mirror

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★ We know that....

$\boxed{ \purple{:\longmapsto}\rm \dfrac{1}{f} \: = \: \dfrac{1}{v} \: - \: \dfrac{1}{u}}$

( Mirror Formula )

Put the known values in above equation :

$\purple{: \implies}\rm\dfrac{1}{-15} \: = \: \dfrac{1}{v} \: - \: \dfrac{1}{-20}$

$\: \: \: \: \: \: \: \: \: \purple{: \implies}\rm\dfrac{-1}{15}+ \dfrac{(-1)}{20} = \dfrac{1}{v}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{: \implies} \rm \dfrac{1}{v} = \dfrac{ - 4 + 3}{60}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \purple{: \implies} \rm \dfrac{1}{v} = \dfrac{1}{60}$

$\: \: \: \: \: \purple{: \: \longmapsto} \boxed{\boxed{\rm \: v = 60 \: cm}}$

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★ Now...

$\: \: \: \: \: \boxed{ \rm*\:m\:=\:\dfrac{-v}{u}\:=\:\dfrac{h}{h'}}$

$\: \: \: \: \: \: \: \red{: \implies}\rm\dfrac{-60}{20} = \dfrac{h}{5}$

$\: \: \: \: \: \: \: \red{ : \implies} \rm\dfrac{h}{5} = - 3$

$\: \: \: \: \: \: \: \: \: \: \: \: \dag \: \boxed{ \rm\purple{\underline{h \:=\: - 15 \:cm}}} \: \dag$