Question

A 5cm tall object is placed at adistance of 30cm from a convex mirror of focal length 15cm find the position size and nature of the image formed​

Answer 1

Given :

In convex mirror,

Object height = 5 cm

Object distance = - 30cm

Focal length = 25 cm

To Find :

The position, size and the nature of the image formed.

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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 30} = \dfrac{1}{25}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{ 30} = \dfrac{1}{25}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{25} + \dfrac{1}{30}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{6 + 5}{150}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{11}{150}$

$\dashrightarrow\sf v = \dfrac{150}{11}$

$\dashrightarrow\sf v = 13.6 \: cm$

Thus, the position of the image is 13.6 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,

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

$\dashrightarrow\sf - \dfrac{13.6}{30} = \dfrac{h'}{5}$

$\dashrightarrow\sf h' = - \dfrac{13.6 \times 5}{30}$

$\dashrightarrow\sf h' = - \dfrac{68}{30}$

$\dashrightarrow\sf h' = - 2.2 \: cm$

Thus, the size of the image is -2.2 cm.

Nature of the image :

• The image is virtual and erect.
• The image is diminished.