Question

An object of size 5cm is placed at a distance 25cm from a convex mirror of focal length 75cm.Find
position, size and nature of the image formed.
​

Answer 1

Given:-

• Height of object ,ho = 5cm

• Object distance ,u = -25 cm

• Focal length ,f = 75 cm

• Types of Mirror = Convex Mirror

To Find:-

• Position , Size and Nature of the image formed

Solution:-

Firstly we calculate the image distance.

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/(-25) = 1/75

→ 1/v -1/25 = 1/75

→ 1/v = 1/75 +1/25

→ 1/v =1+3 /75

→ 1/v = 4/75

→ v = 75/4

→ v = 18.75 cm

Therefore, the image distance is 18.75 cm.

Now, Calculating the Height of image. Using Magnification Formula

• m = hi/ho = -v/u

Substitute the value we get

→ hi/5 = -75/5/(-25)

→ hi/5 = 75/5/25

→ hi/5 = 3/5

→ hi = 15/5

→ hi = 3 cm

Therefore,the image height is 3 cm and the image formed is smaller than the object .

Nature of Image :- image formed is Virtual and erect.

Answer 2

Answer:-

➪ The position of image, v = $\boxed{\tt{\red{+18.75\:cm}}}$

➪ Size of the image, $\sf{h_i}$ = $\boxed{\tt{\red{3.75\:cm}}}$

➪ Nature of the image - Virtual and erect.

$\:$

Given :-

• Height of object ($\sf{h_o}$) = 5 cm

• Object distance (u) = - 25 cm

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$(∵ Left to pole)

• Focal length (f) = + 75 cm

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$(∵ Right to pole)

• It is a convex mirror.

$\:$

To find :-

• The position of image, v = ?

• Size of the image, $\sf{h_i}$ = ?

• Nature of the image.

$\:$

Formula used :-

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

• $\boxed{\sf{Magnification= \dfrac{h_i}{h_o}= -\dfrac{v}{u}}}$

$\:$

Solution :-

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

Substituting the known values in the above equation we get,

$\sf{⇒\dfrac{1}{75} = -\dfrac{1}{25} +\dfrac{1}{v}}$

$\sf{⇒\dfrac{1}{v} = \dfrac{1}{75} +\dfrac{1}{25}}$

$\sf{⇒\dfrac{1}{v} = \dfrac{1}{75} +\dfrac{1}{25}}$

$\sf{⇒\dfrac{1}{v} = \dfrac{1+3}{75}}$

$\sf{⇒v = }$ $\sf{\red{+ 18.75\:cm}}$

$\:$

$\sf{⇒Magnification= \dfrac{h_i}{h_o}= -\dfrac{v}{u}}$

From this,

$\sf{⇒ \dfrac{h_i}{h_o}= -\dfrac{v}{u}}$

$\sf{⇒ \dfrac{h_i}{5}= -\dfrac{18.75}{(-25)}}$

$\sf{⇒ h_i= \dfrac{93.75}{25}}$

$\sf{⇒ h_i= }$$\sf{\red{3.75\:cm}}$