Question

An object of height 10 cm is placed 20 cm away from a convex mirror. The image is formed 25 cm away from the object. Find the height of the object

Answer 1

Given :

In convex mirror,

Height of the object : 10 cm

Object distance : 20 cm

Image distance : 25 cm

To find :

The height of the object.

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,

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

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,

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

$\dashrightarrow\sf \dfrac{h'}{10} = - \dfrac{25}{ - 20}$

$\dashrightarrow\sf h' = - \dfrac{25 \times 10}{ - 20}$

$\dashrightarrow\sf h' = - \dfrac{250}{ - 20}$

$\dashrightarrow \underline{\boxed{\sf h' = 12.5 \: cm}}$

Thus, the height of the image is 12.5 cm.

Answer 2

Answer:-

• Height of the object=h=10cm
• Object distance=-20cm=u
• Image distance=25cm=v
• Image height=h'=?

We know that

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

$\\ \qquad\quad\sf\Rrightarrow \dfrac{h'}{h}=-\dfrac{v}{u}$

$\\ \qquad\quad\sf\Rrightarrow \dfrac{h'}{10}=-\dfrac{25}{-20}$

$\\ \qquad\quad\sf\Rrightarrow \dfrac{h'}{10}=\dfrac{25}{20}$

$\\ \qquad\quad\sf\Rrightarrow h'=\dfrac{25\times 10}{20}$

$\\ \qquad\quad\sf\Rrightarrow h'=\dfrac{250}{20}$

$\\ \qquad\quad\sf\Rrightarrow h'=12.5cm$