Question

a concave mirro produces a real image of 1cm tall of an image of 2.5cm tall placed at 5cm from the mirror. Find the position of the image and the focal length of the mirror

Answer 1

Given :

In concave mirror,

Height of the image : 1 cm

Height of the object : 2.5 cm

Object distance : 5 cm

To find :

The position of the image and the focal length of the image.

Solution :

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,

$\dashrightarrow\sf \dfrac{1}{2.5} = - \dfrac{ v}{5}$

$\dashrightarrow\sf \dfrac{5}{2.5} = - v$

$\dashrightarrow\sf v = - 2 \: cm$

Thus, the position of the image is 2 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,

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

$\dashrightarrow\sf \dfrac{1}{ - 2} + \dfrac{1}{ - 5} = \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{1}{ - 2} - \dfrac{1}{ 5} = \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{ - 5 - 2}{ 10} = \dfrac{1}{f}$

$\dashrightarrow\sf \dfrac{ - 7}{ 10} = \dfrac{1}{f}$

$\dashrightarrow\sf f= - \dfrac{10}{7}$

$\dashrightarrow\sf f= - 1.4 \: cm$

Thus, the focal length of the image is 1.4 cm.