Question

An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.

Answer 1

$\textbf{Answer is in Attachment !}$

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Given :-

Focal length

= R/2

= 30/2

= 15 cm

Also,

u = - 20 cm (Object distance)

h₁ = 5 cm (Height of object)

Now,

Using Mirror Formula,

1/v + 1/u = 1/f

1/v + 1/- 20 = 1/15

1/v = 1/15 + 1/20

1/v = 4 + 3/60

1/v = 60/7

v = 8.57 cm

Therefore,

Image distance = 8.57 cm.

Now,

m = h₂/h₁ (Magnification)

m = -v/u

m = h₂/h₁

m = -v/u

h₂/5 = 8.57/20

h₂ = 8.57 × 5/20

h₂ = 2.175 cm.

Therefore,

Height of image = 2.175 cm

Image = virtual and erect