Science, asked by Disha976, 6 months ago

Find the size, nature and position of the image formed when an object of size 1cm is placed at a distance of 15cm from a concave mirror of focal length 10cm.​

Answers

Answered by Yuseong
14

 {\underline {\underline {\huge {\bf { Provided \: Question: } }}}}

Find the size, nature and position of the image formed when an object of size 1 cm is placed at a distance of 15cm from a concave mirror of focal length 10 cm.

 {\underline {\underline {\huge {\bf { Required \: Answers: } }}}}

Position of the image:

To find the position of the image, we'll use here mirror formula-

Given:

  • » u ( Object distance ) = -15cm

( displacement taken left side of the mirror is taken always negatively)

  • » f ( focal length ) = -10cm

( As it is concave mirror )

To find:

  • v ( image distance i.e position of the image)

Calculation:

We know,

 \large \sf { \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}

⠀⠀

Putting the values -

 \sf { : \implies \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}

 \sf { : \implies \dfrac{1}{v} + \dfrac{1}{-15} = \dfrac{1}{-10}}

 \sf { : \implies \dfrac{1}{v} - \dfrac{1}{15} = - \dfrac{1}{10}}

\sf { : \implies \dfrac{1}{v} = - \dfrac{1}{10}+ \dfrac{1}{15} }

 \sf { : \implies \dfrac{1}{v} = \dfrac{-3+2}{30} }

\sf { : \implies \dfrac{1}{v} = - \dfrac{1}{30} }

\sf { : \implies v = - 30 \: cm}

Therefore, the position of the image is -30cm i.e 30cm to the left side of the mirror. \purple {\bigstar}

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Nature of the image:

As the image is formed to the left side of the mirror, so it will be real and inverted image.★

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Size of the image:

To find the nature of the image we need to calculate its magnification first.

 \large \sf {m = - \dfrac{v}{u}}

Where,

  • m = magnification = ?
  • v = image distance = -30cm
  • u = object distance = - 15cm

Put the values-

 \sf {: \implies m = - \dfrac{\cancel {-} 30}{\cancel {-} 15}}

\sf {: \implies m = \cancel { - \dfrac{30}{ 15}} }

 \sf {: \implies m = - 2}

Therefore the size of the image is 2cm.( – ) sign indicates that the image is formed below the principal axis which is real and inverted.  \purple{ \bigstar}

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