Question

A object placed at a distance of 6cm in front of a concave mirror whose focal length is 12cm. Find the position,nature,and size of the image

Answer 1

Given :

After sign convention

• focal length,f =-12cm
• u = -6cm

To find :

• The position of image
• Nature and size of the image

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

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

Solution :

By using Mirror Formula

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

$\sf\implies\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}$

Now put rhe given values

$\sf\implies\dfrac{1}{v}=\dfrac{-1}{12}-\dfrac{(-1)}{6}$

$\sf\implies\dfrac{1}{v}=\dfrac{-1}{12}+\dfrac{1}{6}$

$\sf\implies\dfrac{1}{v}=\dfrac{-1+2}{12}$

$\sf\implies\:v=12cm$

since, object is placed between P and F , we get virtual and erect.

We know the magnification :

$\sf\:m=\dfrac{-v}{u}$

$\sf\implies\:m=\dfrac{(-12)}{(-6)}$]

$\sf\:m=2$

Now ,we assume that height of object is x cm.

We know that

$\sf\:m=\dfrac{h_{i}}{h_{o}}$

$\sf\implies\:2=\dfrac{h_{i}}{x}$

$\sf\implies\:h_{i}=2x\:cm$

Conclusion:

• Postion of image ,v =12cm
• Nature of image : virtual and erect
• Height of image ,h = 2x =2×height of object

Answer 2

Explanation:

Given :

After sign convention

focal length,f =-12cm

u = -6cm

To find :

The position of image

Nature and size of the image

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

MirrorFormula

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

f

1

=

v

1

+

u

1

Solution :

By using Mirror Formula

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

f

1

=

v

1

+

u

1

\sf\implies\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}⟹

v

1

=

f

1

−

u

1

Now put rhe given values

\sf\implies\dfrac{1}{v}=\dfrac{-1}{12}-\dfrac{(-1)}{6}⟹

v

1

=

12

−1

−

6

(−1)

\sf\implies\dfrac{1}{v}=\dfrac{-1}{12}+\dfrac{1}{6}⟹

v

1

=

12

−1

+

6

1

\sf\implies\dfrac{1}{v}=\dfrac{-1+2}{12}⟹

v

1

=

12

−1+2

\sf\implies\:v=12cm⟹v=12cm

since, object is placed between P and F , we get virtual and erect.

We know the magnification :

\sf\:m=\dfrac{-v}{u}m=

u

−v

\sf\implies\:m=\dfrac{(-12)}{(-6)}⟹m=

(−6)

(−12)

\sf\:m=2m=2

Now ,we assume that height of object is x cm.

We know that

\sf\:m=\dfrac{h_{i}}{h_{o}}m=

h

o

h

i

\sf\implies\:2=\dfrac{h_{i}}{x}⟹2=

x

h

i

\sf\implies\:h_{i}=2x\:cm⟹h

i

=2xcm

Conclusion:

Postion of image ,v =12cm

Nature of image : virtual and erect

Height of image ,h = 2x =2×height of object