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 object

Answer 1

Given:

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.

To find:

Image height?

Calculation:

In case of mirrors, we know that ratio of image distance and object distance is equal to the ratio of image height and object height.

$\rm \: \dfrac{ h_{i}}{ h_{o} } = - \dfrac{v}{u}$

$\rm \implies\: \dfrac{ h_{i}}{10} = - \dfrac{25}{( - 20)}$

$\rm \implies\: \dfrac{ h_{i}}{10} = \dfrac{25}{20}$

$\rm \implies\: \dfrac{ h_{i}}{10} = \dfrac{5}{4}$

$\rm \implies\: h_{i} = 12.5 \: cm$

So, image height will be 12.5 cm, and the image is erect and virtual.

Answer 2

SOLUTION

GIVEN

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

TO DETERMINE

The height of the image

EVALUATION

Here it is given that an object of height 10 cm is placed 20 cm away from a convex mirror.

The image is formed 25 cm away from object

h = Height of the object = 10 cm

v = Image distance = 25 cm

u = Object distance = - 20 cm

h' = Height of the image = ?

We know that

$\displaystyle \sf{Linear \: magnification = m = \frac{Height \: of \: the \: image}{Height \: of \: the \: object} = \frac{h'}{h} }$

Also linear magnification is related to Image distance and Object distance as below

$\displaystyle \sf{Linear \: magnification = m = - \frac{Image \: distance}{Object \: distance} = - \frac{v}{u} }$

Thus we get

$\displaystyle \sf{ m = \frac{h'}{h} = - \frac{v}{u} }$

Putting the values we get

$\displaystyle \sf{ \frac{h'}{10} = - \frac{25}{ - 20} }$

$\displaystyle \sf{ \implies \: h' = \frac{25}{ 20} \times 10 }$

$\displaystyle \sf{ \implies \: h' = \frac{25}{ 2} }$

$\displaystyle \sf{ \implies \: h' = 12.5}$

FINAL ANSWER

The height of the image = 12.5 cm

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