Question

An object of height 5 cm is held 20 cm away from converging mirror of focal length 10 cm

find the position , nature and size of the image formed .
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Answer 1

Explanation:

converging mirror means concave mirror

sine convemtions f(focal length -ve) u(object distance -ve) v ( image distance - ve) exept one case i.e between focal length and principle axis you should know if you are in 10th

height(h) = 5 cm

u. = -20 cm see sign convemtions

v. =. ?

f. =. - 10 cm

1/u+1/v=1/f (mirror formulae)

-1/20+1/v= -1/10

1/v= -1/10+1/20

1/v= -1/20

v= - 20 cm

position is - 20 cm

virtual and erect as it is behind the mirror

size of image = h'

to find -:

-v/u = h'/h

h is given

-(-20)/(-10)=h'/5

-2×5=h'

-10 = h'

height of image = 10 cm as height can never be negative

Hope it helps it is very important ch for cbse class 10 exams study well bro....☺️✌️

Answer 2

Given :

Height of object  h₀ = 5cm

Distance of object u = - 20cm

Focal length f = 10 cm

Type of mirror =  Concave mirror

                       

To find :

Position of image v

Nature of image

Size of image / Height of image $h_{i}$

Solution :

We know ,

   $\frac{1}{f}\ = \frac{1}{v} \ + \ \frac{1}{u}$

⇒ $\frac{1} {10} \ = \frac{ 1 }{ v } \ - \ \frac{ (- 1) }{20}$

⇒ $\frac{1}{v} = \frac{1}{10} - \frac{1 }{20}$

⇒ $\frac{1}{v} = \frac{ 2 - 1 }{ 20 } = \frac{1 }{20}$

⇒ $\frac{1}{v} = \frac{1 } {20}$

║▣  v = 20 cm ▣║

Position of image : Formed at 20 cm from the lens and on the other side of the lens with  respect to the object.

Nature of image : Real and inverted.

We know

Magnification (M) =   $\frac{h_{i} } {h_{o} } = \frac{ - v}{u}$

⇒$\frac{20}{-20} = - 1 = \frac{h_i}{h_o}$   ( since, h₀ = 5cm )

∴ $h_{i} = - 5cm$

Height/ Size of image : Height of image or size of image is same as the height of object or size of object numerically.