Question

A 9 cm needle is placed 24 cm away from a
convex mirror of focal length 30 cm. The
height of image is-
04.5 cm
o 18 cm
45 cm
5 cm​

Answer 1

Given :

In Convex mirror,

Object height = 9cm

Object distance = - 24cm

Focal length = 30cm

To find :

Height of the image

Solution :

Firstly we have to find position of image 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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{( - 24)} = \dfrac{1}{30}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{24} = \dfrac{1}{30}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{30} + \dfrac{1}{24}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{4 + 5}{120}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{9}{120}$

$\dashrightarrow\sf v = \dfrac{120}{9}$

$\dashrightarrow\sf v = 13.3 \: cm$

Now, using magnification formula that is,

» The Magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign and is also equal to the ratio of height of the image to the height of the object .i.e.,

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

where,

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

By substituting all the given values in the formula,

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

$\dashrightarrow\sf - \dfrac{ \dfrac{120}{9} }{24} = \dfrac{h'}{9}$

$\dashrightarrow\sf h' = - \dfrac{120 \times 9}{24 \times 9}$

$\dashrightarrow\sf h' = - \dfrac{120}{24}$

$\dashrightarrow\sf h' = - 5 \: cm$

Thus, the height of the image is 5cm.

Answer 2

Explanation:

Given :

In Convex mirror,

Object height = 9cm

Object distance = - 24cm

Focal length = 30cm

To find :

Height of the image

Solution :

Firstly we have to find position of image 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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{( - 24)} = \dfrac{1}{30}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{24} = \dfrac{1}{30}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{30} + \dfrac{1}{24}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{4 + 5}{120}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{9}{120}$

$\dashrightarrow\sf v = \dfrac{120}{9}$

$\dashrightarrow\sf v = 13.3 \: cm$

Now, using magnification formula that is,

» The Magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign and is also equal to the ratio of height of the image to the height of the object .i.e.,

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

where,

v denotes image distance

u denotes object distance

h' denotes image height

h denotes object height

By substituting all the given values in the formula,

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

$\dashrightarrow\sf - \dfrac{ \dfrac{120}{9} }{24} = \dfrac{h'}{9}$

$\dashrightarrow\sf h' = - \dfrac{120 \times 9}{24 \times 9}$

$\dashrightarrow\sf h' = - \dfrac{120}{24}$

$\dashrightarrow\sf h' = - 5 \: cm$

Thus, the height of the image is 5cm.