Physics, asked by raiutkarsh28721, 4 months ago

An object of height 5 cm is placed 20cm in front of a concave mirror of focal length 15 cm find the image distance and calculate the size of the image formed

Answers

Answered by BrainlyTwinklingstar
16

Given :

Object height = 5cm

Object distance = -20cm

Focal length = -15cm

Mirror : concave mirror

Note : In concave mirror focal length is always negative.

To find :

The image distance a nd the size of the image formed.

Solution :

using mirror formula that is,

» A formula which gives the relationship between image distance, object distance and focal length of a sperical mirror is known as the mirror formula .i.e.,

\boxed{ \bf \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }

where,

  • v denotes Image distance
  • u denotes object distance
  • f denotes focal length

now, substituting all the given values,

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

\leadsto{ \sf \dfrac{1}{v} + \dfrac{1}{( - 20)} = \dfrac{1}{( - 15)} }

\leadsto{ \sf \dfrac{1}{v}  -  \dfrac{1}{20} = -  \dfrac{1}{15} }

\leadsto{ \sf \dfrac{1}{v}  =  -  \dfrac{1}{15}  +  \dfrac{1}{20} }

\leadsto{ \sf \dfrac{1}{v}  =  \dfrac{ - 4 + 3}{60}  }

\leadsto{ \sf \dfrac{1}{v}  =  -  \dfrac{ 1}{60}  }

\leadsto{ \sf v =  - 60 \: cm }

thus, the image distance is -60cm.

» The Magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign and is also equal to the ratio of height of the image to the height of the object .i.e.,

\boxed{ \bf m = - \dfrac{v}{u} = \dfrac{h'}{h}}

where,

  • v denotes image distance
  • u denotes object distance
  • h' denotes image height
  • h denotes object height

Now, substituting all the given values in the formula,

\leadsto{ \sf -\dfrac{v}{u} =  \dfrac{h'}{h} }

\leadsto{ \sf -\dfrac{( - 60)}{( - 20)} =  \dfrac{h'}{5} }

\leadsto{ \sf -3 =  \dfrac{h'}{5} }

\leadsto{ \sf  h' =  - 3 \times 5 }

\leadsto{ \sf  h' =  - 15  \: cm}

thus, the size of the image is -15cm.

The minus signhere shows that the image is formed below the principle axis. This is the image is real and inverted.

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