Question

a object 2 cm tall placed at distance of 30cm from converging mirror has radius of carvature of 20cm find position size and nature

Answer 1

Answer:

According to the question:

Object distance, u=−30 cm

Focal length, f=−15 cm

Let the Image distance be v.

By mirror formula:

v

1

+

u

1

=

f

1

[4pt]

⇒

v

1

+

−30 cm

1

=

−15 cm

1

[4pt]

⇒

v

1

=−

−30 cm

1

+

−15 cm

1

[4pt]

⇒

v

1

=

30 cm

1−2

[4pt]

⇒

v

1

=−

30 cm

1

[4pt]

∴v=−30 cm

Thus, screen should be placed 30 cm in front of the mirror (Centre of curvature) to obtain the real image.

Now,

Height of object, h

1

=2 cm

Magnification, m=

h

1

h

2

=−

u

v

Putting values of v and u:

Magnification m=

2 cm

h

2

=−

−30 cm

−30 cm

⇒

2 cm

h

2

=−1

[4pt];

⇒h

2

=−1×2 cm=−2 cm

Thus, the height of the image is 2 cm and the negative sign means the image is inverted.

Thus real, inverted image of size same as that of object is formed.

The diagram shows image formation.

Answer 2

Given :

In converging mirror,

Height of the object : 2 cm

Object distance : 30 cm

Radius of curvature : 20 cm

To find :

The position, size and nature of the image.

Solution :

First we have to find focal length we know that,

» For a spherical mirror having small aperture, the principle focus lies exactly mid way between the pole and centre of curvature. So, the focal length of a spherical mirror is equal to the half of its radius of curvature.

if f is the focal length of a mirror and R is its radius of curvature, then f = R/2

by substituting the given values in the formula,

$\dashrightarrow \sf f = \dfrac{R}{2}$

$\dashrightarrow \sf f = \dfrac{20}{2}$

$\dashrightarrow \sf f = 10 \: cm$

Now, 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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 30} = \dfrac{1}{ - 10}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{30} = \dfrac{1}{ - 10}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{ - 10} + \dfrac{1}{30}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 3 + 1}{30}$

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

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

$\dashrightarrow\sf v = - 15 \: cm$

Thus, the position of the image is 15 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,

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

where,

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

By substituting all the given values in the formula,

$\dashrightarrow\sf \dfrac{h'}{2} = - \dfrac{ - 15}{ - 30}$

$\dashrightarrow\sf \dfrac{h'}{2} = - \dfrac{1}{2}$

$\dashrightarrow\sf h' = - 1 \: cm$

Thus, the height of the image is 1 cm

Nature of the image :

• The image is real and inverted.
• The image is diminished.