Question

A 5cm tall object is placed 25cm away from the focus of concave mirror of facial length 20cm. what will be the distance and size of the image formed? also tell about nature and position of the image​

Answer 1

Given :

In concave mirror,

Height of the object : 5 cm

Object distance : - 25 cm

Focal length : - 20 cm

To find :

The position, size and nature of the image.

Solution :

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,

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

$\dashrightarrow\sf \dfrac{1}{v} + \dfrac{1}{ - 25} = \dfrac{1}{ - 20}$

$\dashrightarrow\sf \dfrac{1}{v} - \dfrac{1}{25} = \dfrac{1}{ - 20}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{1}{ - 20} + \dfrac{1}{25}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 5 + 4}{100}$

$\dashrightarrow\sf \dfrac{1}{v} = \dfrac{ - 1}{100}$

$\dashrightarrow\sf v = - 100 \: cm$

Thus, the position of image is -100 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,

$\dashrightarrow\sf \dfrac{h'}{5} = - \dfrac{ - 100}{ - 25}$

$\dashrightarrow\sf \dfrac{h'}{5} = - 4$

$\dashrightarrow\sf h' = - 20 \: cm$

Thus, the size of the image is-20 cm

Nature of the image :

• The image is real and inverted.
• The image is highly enlarged.