Question

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​

Answer 1

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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Answer 2

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]