Question

An object 5cm high is placed at a distance of 15cm in front of a converging mirror of focal length 30cm. Calculate the image distance and size of the image formed.​

Answer 1

Given :

In converging mirror (concave mirror),

Height of object : 5 cm.

Object distance : 15 cm.

Focal length : 30 cm.

To find :

The image distance and the size of the image formed.

Solution :

Using mirror formula that is,

» A formula which gives the relationship between image distance, object distance and focal length of a sperical mirror is known as the mirror formula .i.e.,

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

where,

• v denotes Image distance
• u denotes object distance
• f denotes focal length

By substituting all the given values in the formula,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 15} = \dfrac{1}{ - 30}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{15} = \dfrac{1}{ - 30}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{ - 30} + \dfrac{1}{15}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 1 + 2}{30}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{30}$

$\dashrightarrow\sf v = 30 \: cm$

Thus, the position of the image is 30 cm.

We know that,

» The linear magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign and it is equal to the ratio of image height and object height. that is,

$\boxed{\bf m = \dfrac{h'}{h} = - \dfrac{v}{u}}$

where,

• h' denotes height of image
• h denotes object height
• v denotes image distance
• u denotes object distance

By substituting all the given values in the formula,

$\dashrightarrow\sf \dfrac{h'}{h} = - \dfrac{v}{u}$

$\dashrightarrow\sf \dfrac{h'}{5} = - \dfrac{ - 30}{ - 15}$

$\dashrightarrow\sf \dfrac{h'}{5} = - 2$

$\dashrightarrow\sf h'= - 2 \times 5$

$\dashrightarrow\sf h'= -10 \: cm$

Thus, the height of the image is 10 cm.

Answer 2

Answer:

Given :

In converging mirror (concave mirror),

Height of object : 5 cm.

Object distance : 15 cm.

Focal length : 30 cm.

To find :

The image distance and the size of the image formed.

Solution :

Using mirror formula that is,

» A formula which gives the relationship between image distance, object distance and focal length of a sperical mirror is known as the mirror formula .i.e.,

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

v

1

+

u

1

=

f

1

where,

v denotes Image distance

u denotes object distance

f denotes focal length

By substituting all the given values in the formula,

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

v

1

+

u

1

=

f

1

\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 15} = \dfrac{1}{ - 30}⇢

v

1

+

−15

1

=

−30

1

\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{15} = \dfrac{1}{ - 30}⇢

v

1

−

15

1

=

−30

1

\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{ - 30} + \dfrac{1}{15}⇢

v

1

=

−30

1

+

15

1

\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 1 + 2}{30}⇢

v

1

=

30

−1+2

\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{30}⇢

v

1

=

30

1

\dashrightarrow\sf v = 30 \: cm⇢v=30cm

Thus, the position of the image is 30 cm.

We know that,

» The linear magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign and it is equal to the ratio of image height and object height. that is,

\boxed{\bf m = \dfrac{h'}{h} = - \dfrac{v}{u}}

m=

h

h

′

=−

u

v

where,

h' denotes height of image

h denotes object height

v denotes image distance

u denotes object distance

By substituting all the given values in the formula,

\dashrightarrow\sf \dfrac{h'}{h} = - \dfrac{v}{u}⇢

h

h

′

=−

u

v

\dashrightarrow\sf \dfrac{h'}{5} = - \dfrac{ - 30}{ - 15}⇢

5

h

′

=−

−15

−30

\dashrightarrow\sf \dfrac{h'}{5} = - 2⇢

5

h

′

=−2

\dashrightarrow\sf h'= - 2 \times 5⇢h

′

=−2×5

\dashrightarrow\sf h'= -10 \: cm⇢h

′

=−10cm

Thus, the height of the image is 10 cm.