Physics, asked by rajgurukumar, 7 months ago

An object 4.0 cm in size is placed at 25.0 cm in front of concave mirror of focal length 15.0 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and size of the image

1) v=-37.5 cm, Image is inverted and enlarge

2)v=37.5 cm, Image is erect and diminished

3)v=-12.5 cm, Image is inverted and enlarge

4)v=12.5 cm, Image is erect and diminished

Answers

Answered by Anonymous
9

GIVEN :

  • Size of an object is 4.0 cm.
  • Object is placed at 25.0 cm in front of a concave mirror.
  • Focal length = 15.0 cm.

TO FIND :

  • At what distance from the mirror should a screen be placed in order to obtain a sharp image ?
  • Nature and size of the image.

FORMULAS USED :

  • Distance of the image = \sf \dfrac {1}{v} \: + \: \dfrac {1}{u} \: = \: \dfrac {1}{f}
  • Size of the image = \sf \dfrac {-v}{u}

SOLUTION :

To find the distance of the image,

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

\implies \sf \dfrac {1}{v} \: + \: \dfrac {1}{-25} \: = \: \dfrac {1}{-15}

\implies \sf \dfrac {1}{v} \: = \: \dfrac {1}{-15} \: + \: \dfrac {1}{25}

\implies \sf v \: = \: \dfrac {-10 + 6}{150}

\implies \sf v \: = \: \dfrac {-4}{150}

\implies \sf v \: = \: -37.5 \: cm.

\red {\boxed{\sf v \: = \: -37.5 \: cm}}

\therefore The screen should be placed 37.5 cm in front of lens.

Now,

To find the size of the image,

Size of the image = \sf \dfrac {-v}{u}

\implies \sf h \: = \: \dfrac {-v \times h_0}{u}

\implies \sf h \: = \: \dfrac {-(-37.5)\times 4}{-25}

\implies \sf h \: = \: \dfrac {-150}{25}

\implies \sf h \: = \: - 6 \: cm

\red {\boxed{\sf h \: = \: - 6 \: cm}}

\therefore The image will be inverted.

Because the value of h is negative.

Answered by Anonymous
0

1) v=-37.5 cm, Image is Inverted and enlarge...

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