Question

an object 4cm is placed at a distance of 6 cm in front of a concave mirror of focal length 12cm . find the position nature and size of the image formed

Answer 1

HERE IS UR ANSWER

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Given:

after sign conventions:

f=-12cm

u=-6cm

h0=4cm

By mirror formula,

1/f=1/v+1/u

1/v=1/f-1/u

=1/-12 +1/6

=-1/12+1/6

=1/12

v=12cm

since object is placed between P and F , we get virtual and erect

m=-v/u

=-12/-6

=+2

also m=hi/h0

2=hi/4cm

hi=8cm

hence image is magnified , that is 8cm height.

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Answer 2

$\huge{\mathfrak{Question:-}}$

An object 4cm is placed at a distance of 6 cm in front of a concave mirror of focal length 12cm . find the position nature and size of the image formed.

$\huge{\mathfrak{Solution:-}}$

$\huge{\bf{\underline{Given:-}}}$

$\sf{f = -12\;cm}$

$\sf{u = -6\;cm}$

$\sf{h_{o}=4\;cm}$

By mirror formula,

$\sf{\frac{1}{v}+\frac{1}{u}=\frac{1}{f}}$

$\sf{\frac{1}{v}=\frac{1}{f}-\frac{1}{u}}$

$\sf{\frac{1}{v}=-\frac{1}{12}+\frac{1}{6}}$

$\sf{\frac{1}{v}=\frac{1}{12}}$

$\boxed{\sf{v = 12\;cm}}$

Hence the object is behind the mirror i.e. image is virtual and erect.

$\sf{m = \frac{-v}{u}}$

$\sf{m = \frac{-12}{-6}}$

$\sf{m = 2\;cm}$

Also, $\sf{m = \frac{h_{i}}{h_{o}}}$

$\sf{2 = \frac{h_{i}}{4\;cm}}$

$\sf{h_{i} = 8\;cm}$

$\boxed{\sf{Hence\;image\;is\;magnified\;i.e.\;8\;cm\;high.}}$

$\bf{Important\;formulas:-}$

$\sf{1).\;Mirror\;formula\;\;\;\;=\;\;\;\; \frac{1}{f}=\frac{1}{v}+\frac{1}{u}}$

$\sf{2).\; Lens\;formula\;\;\;\;\;\;=\;\;\;\;\; \frac{1}{f}=\frac{1}{v}-\frac{1}{u}}$