Question

a 6 cm tall object is placed perpendicular to the principal axis of a concave mirror of focal length 12cm. the distance of the object from the mirror is 16cm. use mirror formula to determine the position, nature and size of the image formed.​

Answer 1

Given type of mirror = Concave

Therefore, acc. to sign convention,

Object's distance,Focal length and image's distance will be negative.

$mirror \: formula \: = \frac{1}{u} + \frac{1}{v} = \frac{1}{f}$

where, u is the distance of the object from the mirror.

v is the distance of image

f is the focal length.

Given, u = 16cm, here -> -16cm

f = 12, here -> -12cm

$\frac{1}{ - 16} + \frac{1}{v} = \frac{1}{ - 12} \\ \\ \frac{1}{v} = \frac{1}{16} - \frac{1}{12} \\ \\ \frac{1}{v} = \frac{3 - 4}{48} \\ \\ \frac{1}{v} = - \frac{1}{48} \\ \\ v = - 48cm$

So, Position of the image is 48cm away from the mirror.

Size -> Given, size of object or height of object, h = 6cm

$we \: know \: that \: m \: = - \frac{v}{u} = \frac{height \: of \: image (h')}{height \: of \: object (h)} \\ \\ - \frac{ - 48}{ - 16} = \frac{h'}{6} \\ \\ - 3 = \frac{h'}{6} \\ \\ h' = 6 \times - 3 = - 18cm$

Therefore, size of image is -18cm, that means it is 18cm tall but inverted.

Nature -> Negative sing of v shows that image is formed in front of the mirror, enlarged, real and inverted.

Answer 2

Answer:

Given type of mirror = Concave

Therefore, acc. to sign convention,

Object's distance,Focal length and image's distance will be negative.

mirror \: formula \: = \frac{1}{u} + \frac{1}{v} = \frac{1}{f}mirrorformula=

u

1

+

v

1

=

f

1

where, u is the distance of the object from the mirror.

v is the distance of image

f is the focal length.

Given, u = 16cm, here -> -16cm

f = 12, here -> -12cm

\begin{gathered}\frac{1}{ - 16} + \frac{1}{v} = \frac{1}{ - 12} \\ \\ \frac{1}{v} = \frac{1}{16} - \frac{1}{12} \\ \\ \frac{1}{v} = \frac{3 - 4}{48} \\ \\ \frac{1}{v} = - \frac{1}{48} \\ \\ v = - 48cm\end{gathered}

−16

1

+

v

1

=

−12

1

v

1

=

16

1

−

12

1

v

1

=

48

3−4

v

1

=−

48

1

v=−48cm

So, Position of the image is 48cm away from the mirror.

Size -> Given, size of object or height of object, h = 6cm

\begin{gathered}we \: know \: that \: m \: = - \frac{v}{u} = \frac{height \: of \: image (h')}{height \: of \: object (h)} \\ \\ - \frac{ - 48}{ - 16} = \frac{h'}{6} \\ \\ - 3 = \frac{h'}{6} \\ \\ h' = 6 \times - 3 = - 18cm\end{gathered}

weknowthatm=−

u

v

=

heightofobject(h)

heightofimage(h

′

)

−

−16

−48

=

6

h

′

−3=

6

h

′

h

′

=6×−3=−18cm

Therefore, size of image is -18cm, that means it is 18cm tall but inverted.

Nature -> Negative sing of v shows that image is formed in front of the mirror, enlarged, real and inverted