A converging mirror forms a real image of height $4 c m$ of an object of height $1 c m$ placed $20 c m$ away from the mirror. Calculate the image distance. What is the focal length of the mirror.

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Solution

Magnification $m = \frac{h_{I}}{h_{O}}$

$m = \frac{- 4}{1} = - 4$

Here, $u = - 20 c m$
Magnification can also be written as:

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

$- 4 = \frac{- v}{- 20}$

$v = - 80 c m$

Thus image distance is $80 c m$ .
Now, Using mirror formula,
$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$
$\frac{1}{f} = \frac{1}{- 80} + \frac{1}{- 20} = \frac{- 5}{80}$
hence, $f = - 16 c m$

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A converging mirror forms a real image of height 4cm of an object of height 1cm placed 20cm away from the mirror. Calculate the image distance. What is the focal length of the mirror.

Magnification m=hIhO m=−41=−4 Here, u=−20cmMagnification can also be written as: m=−vu −4=−v−20 v=−80cm Thus image distance is 80cm.Now, Using mirror formula, 1f=1v+1u 1f=1−80+1−20=−580hence, f=−16 cm

A converging mirror forms a real image of height $4 c m$ of an object of height $1 c m$ placed $20 c m$ away from the mirror. Calculate the image distance. What is the focal length of the mirror.