Question

what is the distance of an object for a real image formed by a concave mirror at distance of 25 cm and focal length of mirror is 5 cm​

Answer 1

Given :

▪ Distance of image = 25cm

▪ Focal length = 5cm

▪ Type of mirror : concave

To Find :

▪ Distance of object.

Concept :

↗ X-coordinate of centre of curvature and focus of concave mirror are negative and those for convex mirdor are positive. In case of mirroe since light rays reflects back in X-direction, therefore -ve sign of v indicates real image and +ve sign of v indicates virtual image.

❇ Mirror Formula :

$\bigstar\:\underline{\boxed{\bf{\red{\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}}}}}$

Calculation :

$\dashrightarrow\sf\:\dfrac{1}{u}+\dfrac{1}{(-25)}=\dfrac{1}{(-5)}\\ \\ \dashrightarrow\sf\:\dfrac{1}{u}=-\dfrac{1}{5}+\dfrac{1}{25}\\ \\ \dashrightarrow\sf\:\dfrac{1}{u}=\dfrac{-5+1}{25}\\ \\ \dashrightarrow\sf\:\dfrac{1}{u}=\dfrac{-4}{25}\\ \\ \dashrightarrow\underline{\boxed{\bf{\purple{u=-6.25\:cm}}}}\:\orange{\bigstar}$

Answer 2

Given:-

❐ Focal length = 5 cm

❐ Distance of image = 25 cm

❐ Type of mirror = concave (-)

To Find:-

Distance of the object

Concept :-

Always remember, Focus of concave mirror are negative and convex mirror are positive. + ve indicates real image and - ve indicates virtual image.

Formula of Mirror:-

$\large\underline{\underline{\boxed{\mathtt{\color{crimson}{\frac{1}{u} + \frac{1}{v} = \frac{1}{f}}}}}}{\color{pink}{\bigstar}}$

Calculation:-

${\implies{\tt{\frac{1}{u} + \frac{1}{(-25)} = \frac{1}{(-5)} }}}$

${\implies{\tt{\frac{1}{u} = - \frac{1}{5} + \frac{1}{25} }}}$

${\implies{\tt{\frac{1}{u} + \frac{-5+1}{25}}}}$

${\implies{\tt{\frac{1}{u} + \frac{-4}{25} }}}$

$\large\implies{\underline{\boxed{\mathtt{\color{purple}{u = 6.25 cm}}}}}{\color{gold}{\bigstar}}$