Question

.An object 2cm high is placed at a distance of 20cm in front of a concave mirror with the focal length of 30cm.Find the image distance and nature of image​

Answer 1

Given :

Object height, h = 2cm

Object distance, u = - 20cm(to the left of mirror)

focal length, f = - 30cm

Remember : In concave mirror focal length is negative.

To find :

the image distance and nature of image

Solution :

using mirror formula,

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

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

here,

v denotes distance of image from mirror

u denotes distance of object from mirror

f denotes focal length

by substituting all the given values in the formula,

${ \leadsto{ \bf{ \dfrac{1}{v } + \dfrac{1}{ - 20} = \dfrac{1}{ - 30} }}}$

${ \leadsto{ \bf{ \dfrac{1}{v } - \dfrac{1}{20} = - \dfrac{1}{30} }}}$

${ \leadsto{ \bf{ \dfrac{1}{v } = - \dfrac{1}{30} + \dfrac{1}{20} }}}$

${ \leadsto{ \bf{ \dfrac{1}{v } = \dfrac{ - 2 + 3}{60} }}}$

${ \leadsto{ \bf{ \dfrac{1}{v } = \dfrac{1}{60} }}}$

${ \leadsto{ \bf{ v = 60 \: cm}}}$

thus, the image distance of the mirror is 60cm.

Magnification of the mirror :

By using the formula for Magnification,

The Linear magnification produced by a mirror is equal to the ratio of the image distance to the object distance with a minus sign.

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

${\leadsto{\bf{ m = \dfrac{-60}{-20}}}}$

${\leadsto{\bf{ m = 3}}}$

${\leadsto{\boxed{\bf{ m = \dfrac{h'}{h}}}}}$

${\leadsto{\bf{ 3 = \dfrac{h'}{2}}}}$

${\leadsto{\bf{ h' = 3× 2}}}$

${\leadsto{\bf{ h' = 6cm.}}}$

As the image distance is greater than object distance the image is enlarged.

Nature of the image :

• the image is virtual and erect
• the image is enlarged

Answer 2

Answer :

• Image distance of the mirror , v = 60 cm.

• Nature of the image is real , virtual and erect.

Explanation :

Given :

• Height of the object, ho = 2 cm

• Object Distance, u = 20 cm

• Focal length, f = 30 cm

To find :

• Image distance of the mirror , v = ?
• Nature of the image

Knowledge required :

• Mirror formula :

⠀⠀⠀⠀⠀⠀⠀⠀⠀1/f = 1/v + 1/u

Where :

• f = Focal length
• v = Image distance
• u = Object Distance

• Formula for magnification :

⠀⠀⠀⠀⠀⠀⠀⠀⠀m = - v/u

Where :

• v = Image distance
• u = Object Distance

[Note : By sign convention in a concave mirror, the focal length and the object distance is taken as negative]

Solution :

Image distance :

By using the mirror formula and substituting the values in it , we get :

==> 1/f = 1/v + 1/u

==> 1/(-30) = 1/v + 1/(-20)

==> 1/(-30) - 1/(-20) = 1/v

==> (-1)/30 + 1/20 = 1/v

==> (-2 + 3)/60 = 1/v

==> 1/60 = 1/v

==> v = 60

∴ v = 60 cm

Magnification of the mirror :

By using the formula for Magnification and substituting the values in it, we get :

==> m = - v/u

==> m = - 60/(-20)

==> m = - (-3)

==> m = 3

∴ m = 3

Thus the Image is larger than the object z virtual and erect.

Therefore,

• Image distance = 60 cm
• Nature of image = large, virtual and erect.