Question

an object of 5cm is placed 1metre in front of a concave mirror . it has a radius of curvature of 20cm.find the position,size and Nature of the image​

Answer 1

Answer

The position of the image is 11.11cm in front of the mirror

The image size is 0.55cm which is so diminished compare to object size

The nature of the image is real and inverted

$\bf \large \underline{Given : }$

• An object of 5cm is placed 1metre in front of a concave mirror
• The concave mirror has a radius of curvature 20cm

$\bf \large \underline{To \: Find : }$

• The position , size and nature of the image

$\bf \large \underline{Solution : }$

Given,

object size , h₁ = 5cm

Object distance, u = -1m = -100cm

radius of curvature , r = -20cm

focal length , f = -10cm

Therefore , using mirror formula we have ,

$\sf \implies \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} \\\\ \sf\implies \dfrac{1}{v} = \dfrac{1}{f} - \dfrac{1}{u} \\\\ \sf\implies \dfrac{1}{v} = \dfrac{1}{-10}-\dfrac{1}{-100} \\\\ \sf\implies \dfrac{1}{v} = \dfrac{-10+1}{100} \\\\ \sf\implies \dfrac{1}{v} = -\dfrac{9}{100} \\\\ \sf\implies v = -\dfrac{100}{9} \\\\ \sf\implies v = -11.11$

Therefore , image distance is -11.11cm i.e. the image will form 11.11cm in front of the mirror

Now using magnification ,

$\sf m = \dfrac{h_{2}}{h_{1}} = -\dfrac{v}{u} \\\\ \sf\implies \dfrac{h_{2}}{h_{1}} = -\dfrac{v}{u} \\\\ \sf\implies h_{2} = -\dfrac{v}{u}\times h_{1} \\\\ \sf\implies h_{2} = -\dfrac{-11.11}{-100}\times 5 \\\\\ \sf\implies h_{2} = -0.555$

Thus, the size of the image is -0.55cm , the negative sign indicates that image formed is inverted and real . Moreover , it has been diminished.

Answer 2

ur answer is attached above ...

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