Question

An object when placed at a distance of 15cm in front of a concave mirror produces a real image at a distance 30cm in front of the mirror Calculate the focal length of the mirror.​

Answer 1

Given :-

• Object distance = u = -15 cm
• Image distance = v = -30 cm

To find :-

• Focal length = f = ?

Solution :-

Since it is a mirror, we apply mirror formula.

$\boxed{ \bf{\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} }}$

$\implies \sf{ \dfrac{1}{ - 30} + \dfrac{1}{ - 15} = \dfrac{1}{f} }$

$\implies \: \sf{ \dfrac{1}{f} = \dfrac{ - 1 - 2}{30} }$

$\implies \: \sf{ \dfrac{1}{f} = \dfrac{ - 3}{30} }$

$\implies \: \sf{ \dfrac{1}{f} = \dfrac{ - 1}{30} }$

$\therefore \: \bf{ \underline{f = - 30 \: cm}}$

So, the focal length is -30 cm

Know more:

• Focal length is the distance between pole and principal focus.
• Object is always placed towards the left of the mirror and so u will always be negative.

Answer 2

Given :-

• Object = u = -15 cm
• Image distance = v = -30 cm

To find :-

• Find the focal length = f = ?

Solution :-

• Since it is a mirror, we apply mirror formula.

$\huge\red\star$$\boxed{\bf{ \frac{1}{v \: } + \frac{1}{u} = \frac{1}{f}}}$$\huge\red\star$

:$\implies$$\large{\bf{ \frac{1}{ - 30} + \frac{1}{ - 15} = \frac{1}{f}}}$

$~~~~~$:$\implies$$\large{\bf{ \frac{1}{f} = \frac{ - 1 \: - 2}{30}}}$

$~~~~~~~~~~$:$\implies$$\large{\bf{ \frac{1}{f} = \frac{ - 3}{30}}}$

$~~~~~~~~~~~~~~~$:$\implies$$\large{\underline{\boxed{\pink{\bf{f~=~-30~cm}}}}}$$\purple\bigstar$

$\large\dag$ Hence,

• The focal length is $\underline{\bf{-30~cm}}$ $\large{\bf\green{✓}}$

$~~~~~~~~~~~~~~$ _____________________

$~~~~~~~~~~~~~~$ ★ More Information :

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• Focal length : Focal length is the distance between pole and principal focus.

• Object is always placed towards the left of the mirror and so you will always be negative.