An object of height 20 cm is kept at a distance of 48 cm in front of a mirror of a focal length of 12 cm. 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.
Answers
Answer:
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 :
The distance of the image from the mirror.
Magnification = ?
FORMULAS USED :
Mirror formula = \sf \dfrac {1}{f} \ = \ \dfrac {1}{v} \ + \ \dfrac {1}{u}
f
1
=
v
1
+
u
1
Magnification = \sf \dfrac {-v}{u}
u
−v
SOLUTION :
To find, the distance of the mirror from the image.
Using mirror formula : \sf \dfrac {1}{f} \ = \ \dfrac {1}{v} \ + \ \dfrac {1}{u}
f
1
=
v
1
+
u
1
\implies \sf \dfrac {1}{v} \ = \ \dfrac {1}{f} \ - \ \dfrac {1}{u}⟹
v
1
=
f
1
−
u
1
\implies \sf \dfrac {1}{v} \ = \ \dfrac {1}{-12} \ - \ \dfrac {1}{-48}⟹
v
1
=
−12
1
−
−48
1
\implies \sf \dfrac {1}{v} \ = \ \dfrac {1}{-12} \ + \ \dfrac {1}{48}⟹
v
1
=
−12
1
+
48
1
\implies \sf \dfrac {1}{v} \ = \ \dfrac {-4+1}{48}⟹
v
1
=
48
−4+1
\implies \sf \dfrac {1}{v} \ = \ \dfrac {-3}{48}⟹
v
1
=
48
−3
\implies \sf \dfrac {1}{v} \ = \ \dfrac {-1}{16}⟹
v
1
=
16
−1
\qquad {\boxed {\sf V \ = \ -16}}
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}Magnification =
u
−v
\implies \sf Magnification \ = \ \dfrac {-(-16)}{48}⟹Magnification =
48
−(−16)
\qquad {\boxed {\sf Magnification \ = \ \dfrac {1}{3}}}
Magnification =
3
1
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