Question

A concave mirror pe produes three times.
enlanged image of an object plaut at 10cm in
front of it. Calculate the focal length of the
mirror​

Answer 1

Given :

• A concave mirror produces three times enlarged image of an object
• And object is placed at 10cm in front of the mirror.

To Find :

Focal length of the mirror .

Theory :

${\red{\boxed{\large{\bold{Mirror\:Formula}}}}}$

$\sf\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

• Magnification

$\sf\:Magnification=\dfrac{h_i}{h_o}=\dfrac{-image\:distance}{object\:distance}$

Solution :

We have

$\sf\:h{i}=3h_{o}$

and $\sf\:u=-10(by\:sign\:convention)$

Magnification

$\sf\:\dfrac{h_i}{h_o}=\dfrac{-(v)}{u}$

Now put the given values

$\sf\:\dfrac{3h_o}{h_o}=\dfrac{-(v)}{-10}$

$\sf\:3=\dfrac{v}{10}$

$\sf\:v=30cm$

Now By using Mirror Formula

$\sf\:\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}$

Put the given values

$\sf\implies\:\dfrac{1}{f}=\dfrac{1}{30}+\dfrac{1}{(-10)}$

$\sf\implies\:\dfrac{1}{f}=\dfrac{1}{30}-\dfrac{1}{10}$

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

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

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

$\sf\implies\:f=-15cm$

Therefore, Focal length of the concave mirror is -15cm.

Answer 2

Given :

A concave mirror produces three times enlarged image of an object

And object is placed at 10cm in front of the mirror.

To Find :

Focal length of the mirror .

Theory :

{\red{\boxed{\large{\bold{Mirror\:Formula}}}}}

MirrorFormula

\sf\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}

f

1

=

v

1

+

u

1

• Magnification

\sf\:Magnification=\dfrac{h_i}{h_o}=\dfrac{-image\:distance}{object\:distance}Magnification=

h

o

h

i

=

objectdistance

−imagedistance

Solution :

We have

\sf\:h{i}=3h_{o}hi=3h

o

and \sf\:u=-10(by\:sign\:convention)u=−10(bysignconvention)

Magnification

\sf\:\dfrac{h_i}{h_o}=\dfrac{-(v)}{u}

h

o

h

i

=

u

−(v)

Now put the given values

\sf\:\dfrac{3h_o}{h_o}=\dfrac{-(v)}{-10}

h

o

3h

o

=

−10

−(v)

\sf\:3=\dfrac{v}{10}3=

10

v

\sf\:v=30cmv=30cm

Now By using Mirror Formula

\sf\:\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}

f

1

=

v

1

+

u

1

Put the given values

\sf\implies\:\dfrac{1}{f}=\dfrac{1}{30}+\dfrac{1}{(-10)}⟹

f

1

=

30

1

+

(−10)

1

\sf\implies\:\dfrac{1}{f}=\dfrac{1}{30}-\dfrac{1}{10}⟹

f

1

=

30

1

−

10

1

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

f

1

=

30

1−3

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

f

1

=

30

−2

\sf\implies\:f=\dfrac{-30}{2}⟹f=

2

−30

\sf\implies\:f=-15cm⟹f=−15cm

Therefore, Focal length of the concave mirror is -15cm.