Question

Find the position, nature and size of the image, When the object of height 3cm is placed at 15cm from length u the convex mirror whose focal length is 10cm.​

Answer 1

Given:-

• Object Distance ,u = -15 cm

• Height of object,ho = 3 cm

• Focal length ,f = 10 cm

To Find:-

• Position , nature & size of the image

Solution:-

Firstly we calculate the image position of the object.

Using mirror formula

❒• 1/v + 1/u = 1/f

where,

v is the Image Position

u is the object distance

f is the focal length

Substitute the value we get

→ 1/v + 1/(-15) = 1/10

→ 1/v -1/15 = 1/10

→ 1/v = 1/10 + 1/15

→ 1/v = 3+2/30

→ 1/v = 5/30

→ 1/v = 1/6

→ v = 6 cm

Therefore , image distance is 6 cm.

Now, Magnification

• m = hi/ho = -v/u

substitute the value we get

→ hi/3 = -6/(-15)

→ hi/3 = 6/15

→ hi/3 = 2/5

→ hi = 6/5 cm

Therefore,the image height is 6/5 cm

Nature of Image :- The image formed is Virtual & erect and smaller than the object .

Answer 2

Given :-

Height = 3 cm

Object distance = -15 cm

Focal length = 10 cm

To Find :-

Position, nature and size

Solution :-

Firstly let's calculate position

As we know that

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

$\sf \: \dfrac{1}{v} + \dfrac{1}{ - 15} = \dfrac{1}{10}$

$\sf \dfrac{1}{v} - \dfrac{1}{15} = \dfrac{1}{10}$

$\sf \: \dfrac{1}{v} = \dfrac{1}{15} + \dfrac{1}{10}$

$\sf \: \dfrac{1}{v} = \dfrac{5}{30}$

$\sf \: \dfrac{1}{v} = \dfrac{6}{1}$

Hence, Position of image is 6 cm

Now,

Let's find size

$\sf \dfrac{hi}{ho} = \dfrac{ - v}{u}$

$\sf \: \dfrac{hi}{3} = \dfrac{ - 6}{-15}$

$\sf \dfrac {hi}{3} = \dfrac{2}{5}$

$\sf \: hi = \dfrac{6}{5} cm$

Hence,

Size of image is 6/5 cm

Nature :-

It is virtual and erect and smaller than object.