A real image of 1/5 the size of the object is formed at a distance of 18 cm from a mirror. The focal length is ____cm.

A real image of

1 / 5

the size of the object is formed at a distance of

18 cm

from a mirror. The focal length is

- 15

cm

.
Given,
Therefore

v = - 18 cm

[Negative: as the image is real]
Size

h_{i} = 1 / 5 h_{o}

Where,

h_{o}

is the size of the object, and

h_{i}

is the size of the image.
Using the magnification(

m

) formula for mirrors,

m = \frac{v}{u}

\Rightarrow \frac{1}{5} = - \frac{18}{u}

\Rightarrow u = - 90 cm

Using mirror formula,

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

Where

f

is the focal length of the lens

\Rightarrow \frac{1}{f} = \frac{1}{- 18} + \frac{1}{- 90}

\Rightarrow \frac{1}{f} = \frac{5}{- 90} - \frac{1}{90}

\Rightarrow \frac{1}{f} = - \frac{6}{90}

\Rightarrow \frac{1}{f} = - \frac{1}{15}

\Rightarrow \frac{1}{f} = - \frac{1}{15}

Therefore

f = - 15 cm .

A real image of $1 / 5$ the size of the object is formed at a distance of $18 cm$ from a mirror. The focal length is ____ $cm$ .