Physics, asked by nongkitlepcha, 5 months ago

An object of height 20 cm is kept at a distance of 48 cm in front of a mirror of focal length 12cm. . If the mirror forms a virtual, diminished image of the object then calculate:

(i) the distance of the image from the mirror.

(ii) its magnification.
its urgent​

Answers

Answered by Anonymous
25

GIVEN :

  • Height of an object = 20 cm
  • Distance = 48 cm
  • Focal length = 12 cm
  • Mirror forms the virtual and diminished image of the object.

TO FIND :

  1. The distance of the image from the mirror.
  2. Magnification = ?

FORMULAS USED :

  • Mirror formula = \sf \dfrac {1}{f} \ = \ \dfrac {1}{v} \ + \ \dfrac {1}{u}
  • Magnification = \sf \dfrac {-v}{u}

SOLUTION :

To find, the distance of the mirror from the image.

Using mirror formula : \sf \dfrac {1}{f} \ = \ \dfrac {1}{v} \ + \ \dfrac {1}{u}

\implies \sf \dfrac {1}{v} \ = \ \dfrac {1}{f} \ - \ \dfrac {1}{u}

\implies \sf \dfrac {1}{v} \ = \ \dfrac {1}{-12} \ - \ \dfrac {1}{-48}

\implies \sf \dfrac {1}{v} \ = \ \dfrac {1}{-12} \ + \ \dfrac {1}{48}

\implies \sf \dfrac {1}{v} \ = \ \dfrac {-4+1}{48}

\implies \sf \dfrac {1}{v} \ = \ \dfrac {-3}{48}

\implies \sf \dfrac {1}{v} \ = \ \dfrac {-1}{16}

\qquad {\boxed {\sf V \ = \ -16}}

\large \therefore The distance of the image from the mirror is -16.

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Now,

To find the magnification = ?

\sf Magnification \ = \ \dfrac {-v}{u}

\implies \sf Magnification \ = \ \dfrac {-(-16)}{48}

\qquad {\boxed {\sf Magnification \ = \ \dfrac {1}{3}}}

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Answered by varshithahansika
3

Answer:

Magnification : 1/3 , Distance of the image from the mirror : -16 ..

Explanation:

(I) Distance of the image from the mirror :

1/v = 1/f -1/u

1/v = 1/-12 - 1/-48

1/v = -4+1/48

1/v = -3/48

1/v = -1/16

[1/v = -16]

(II) Magnification :

-v/u = -(-16)/48

= 16/3

= 1/3

° [Magnification = 1/3]

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