Question

when an object is placed at a distance of 15 cm from a concave mirror ,its image is formed at 10cm infront of the mirror .calculate the focal length of the mirror......
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Answer 1

Explanation:

Here given that

u = - 15 cm

v = - 10 cm

f = ?

By mirror formula

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

$=> \frac{1}{ - 10} + \frac{1}{ - 15} = \frac{1}{f}$

$=> \frac{ - 3 - 2}{30} = \frac{1}{f}$

$=> \frac{ - 5}{30} = \frac{1}{f}$

$=> \frac{1}{ - 6} = \frac{1}{f}$

$=> f = - 6 \: cm$

-ve sign indicates concave mirror whose focal length is 6 cm.

Answer 2

Solution:-

Given:-

$\to\rm \: object \: distance \: (u) = 15cm$

$\rm \to \: image \: distance( v) = 10cm \:$

To find

$\rm \to \: focal \: length(f)$

Formula

$\to \boxed{\rm \: \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u} }$

Now using sign convention

i) object is measured from pole

ii)object distance is opposite to Incident ray so u = -15cm

and image distance is also opposite to Incident ray v = - 10cm

Now using formula

$\rm \: \to \: \dfrac{1}{f} = - \dfrac{1}{ 15} - \dfrac{1}{10}$

$\rm \to \: \dfrac{1}{f} = \dfrac{ - 2 - 3}{30}$

$\to\rm \: \dfrac{1}{f} = \dfrac{ - 5}{30}$

$\rm \to \dfrac{1}{f} = - \dfrac{1}{6}$

$\rm \: f \: = - 6cm$

so focal length is -6 cm