Question

6. A 3cm high object is placed at a distance of 30cm from a concave mirror. A real image is formed 60 cm from the mirror. Calculate the focal length of the mirror and the size of the image. *

(a) 20cm, 6cm

(b) -20cm, -6cm

(c) -16cm, -8cm

(d) -22cm, -7cm

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Answer 1

Given :

Size of object = 3 cm

Distance of object = 30 cm

Distance of image = 60 cm

Type of mirror : concave

To Find :

Focal length of the mirror a d size of the image.

Solution :

❖ Focal length of concave mirror is taken negative and that of convex mirror is taken positive.

• Focal length of the mirror can be calculated by using mirror formula.

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

» u denotes distance of object

» v denotes distance of image

» f denotes focal length

By substituting the given values;

$\sf:\implies\:\dfrac{1}{(-30)}+\dfrac{1}{(-60)}=\dfrac{1}{f}$

$\sf:\implies\:\dfrac{1}{f}=-\dfrac{1}{30}-\dfrac{1}{60}$

$\sf:\implies\:\dfrac{1}{f}=\dfrac{-2-1}{60}$

$\sf:\implies\:f=-\dfrac{60}{3}$

$:\implies\:\underline{\boxed{\bf{\orange{f=-20\:cm}}}}$

♦ Size of the image :

$\sf:\implies\:m=-\dfrac{v}{u}=\dfrac{h'}{h}$

» h' denotes size of the image

» h denotes size of the object

$\sf:\implies\:-\dfrac{(-60)}{(-30)}=\dfrac{h'}{3}$

$\sf:\implies\:h'=3\times (-2)$

$:\implies\:\underline{\boxed{\bf{\gray{h'=-6\:cm}}}}$

∴ (B) is the correct answer!

Answer 2

GIVEN :

• Size of the object = 3cm
• Distance of the object = 30cm
• Real image formation = 60cm

TO FIND :

• Focal length of the mirror = ?
• Size of the image = ?

STEP-BY-STEP INSTRUCTIONS :

Now, we will have to find focal length of the mirror and size of the image = ?

Formula Used :

• Focal length = [ 1/u + 1/v = 1/f ]

→ [ substituting the values as per the formula ]:

→ 1/(-30) + 1/(-60) = 1/f

→ 1/f = -1/30 -1/60

→ 1/f = -2-1/60

→ f = -60/3 = -20

→ Therefore, focal length equals to -20

Formula Used :

• Size = [m = -v/u = h'/h]

[ substituting the values as per the formula ] –

→ m = - (-60)/(-30) = h'/3

→ h' = 3 × (-2) = -6

→ therefore, size=-6