Question

An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius curvature 30 cm. Find the position of the images, it's nature and size.​

Answer 1

Given:

• Object size (hₒ) = 5 cm
• Object distance(u) = 20 cm

[Distances are always measured from the pole of the mirror]

• Radius of curvature= 30 cm [ convex mirror]

Solution:

First, we need to find the focal length of the convex mirror.

we know that

⇒ f = R / 2

⇒ f = 30 / 2

⇒ f = 15 cm

Now,

By applying mirror formula,

$\blacksquare$ 1 / f = 1 / v + 1 / u

⇒ 1 / v = 1 / f - 1 / u

⇒ 1 / v = 1 / 15 - 1 / (-20)

⇒ 1 / v = 1 / 15 + 1 / 20

⇒ 1 / v = 4 + 3 / 60

⇒ 1 / v = 7 / 60

⇒ v = 60/7

⇒ v = 8.57 cm

The image is formed at a distance of 8.57 cm on the other side of the mirror.

Magnification is given by,

$\blacksquare$m = - v / u = hᵢ / hₒ

Hence,

⇒ - v / u = hᵢ / hₒ

⇒ hᵢ = - v x hₒ/ u

⇒ hᵢ = - 8.57 x 5 / - 20

⇒ hᵢ = 42.85 / 20

⇒ hᵢ = 2.14 cm

The height of the image is 2.14 cm.

The image formed is virtual and erect.