Question

An object 5 cm high is placed at a distance 60 cm
in front of a concave mirror of focal length 10 cm.
Find (i) the position and (ii) size, of the image.
Ans. (i) 12 cm in front of the mirror, (ii) 1 cm​

Answer 1

Given :

▪ Distance of object = 60cm

▪ Height of object = 5cm

▪ Focal length of mirror = 10cm

▪ Type of mirror : concave

To Find :

▪ Distance of image.

▪ Size (height) of image.

Formula :

✴ Mirror formula :

$\orange{\bigstar}\:\underline{\boxed{\bf{\purple{\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}}}}}$

✴ Lateral magnification formula :

$\orange{\bigstar}\:\underline{\boxed{\bf{\red{m=\dfrac{h_2}{h_1}=-\dfrac{v}{u}}}}}$

Terms indication :

✒ v denoted distance of image

✒ u denotes distance of object

✒ f denotes focal length of mirror

✒ h$_1$ denotes height of object

✒ h$_2$ denotes height of image

Calculation :

▶ Position of image :

$:\implies\sf\:\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\\ \\ :\implies\sf\:\dfrac{1}{v}+\dfrac{1}{(-60)}=\dfrac{1}{(-10)}\\ \\ :\implies\sf\:\dfrac{1}{v}=-\dfrac{1}{10}+\dfrac{1}{60}\\ \\ :\implies\sf\:\dfrac{1}{v}=\dfrac{-6+1}{60}\\ \\ :\implies\underline{\boxed{\bf{\blue{v=-12cm}}}}\:\orange{\bigstar}$

▶ Size of image :

$\longrightarrow\sf\:\dfrac{h_2}{h_1}=-\dfrac{v}{u}\\ \\ \longrightarrow\sf\:\dfrac{h_2}{5}=-\dfrac{(-12)}{(-60)}\\ \\ \longrightarrow\sf\:h_2=-\dfrac{5}{5}\\ \\ \longrightarrow\underline{\boxed{\bf{\pink{h_2=-1cm}}}}\:\orange{\bigstar}$

____________________________

➡ Nature of image : real & inverted

➡ Size of image : small

Answer 2

(i) Position of image is 12 cm in front of mirror. (ii) Size of image is -1 cm.

Solution

Here we have :-

● Height of object (h_0) = 5 cm.

● Distance of object (u) = -60 cm.

● Focal length (f) = -10 cm.

We will use mirror formula :-

● 1/f = 1/v + 1/u

⇒ 1/-10 = 1/v + 1/(-60)

⇒ 1/-10 = 1/v - 1/60

⇒ 1/v = 1/60 - 1/10

⇒ 1/v = (1 - 6)/60

⇒ 1/v = -5/60

⇒ 1/v = -1/12

⇒ -v = 12

⇒ v = -12 cm

So, position of image is 12 cm in front of mirror.

Now we will use magnification formula :-

● m = h_i/h_o = -v/u

⇒ h_i/5 = -(-12)/-60

⇒ h_i/5 = 12/-60

⇒ h_i/5 = 1/-5

⇒ -5h_i = 5

⇒ h_i = 5/-5

⇒ h_i = -1 cm

So, size of image is -1 cm.