Question

an object is placed at a distance of 40 cm and front of a convex mirror of radius of curvature 20 CM calculate the position and magnification of the object​

Answer 1

Given

• Radius of curvature = 20 cm
• Mirror Used = Convex
• Object Distance = -40 cm

To Find

• Position and magnification of the image

Solution

● Mirror Formula ; 1/v + 1/u = 1/f

● Magnification = -v/u

✭ Focal Length :

→ F = R/2

→ F = 20/2

→ F = 10 cm

✭ Image Distance :

→ 1/v + 1/u = 1/f

→ 1/v + 1/(-40) = 1/10

→ 1/v - 1/40 = 1/10

→ 1/v = 1/10 + 1/40

→ 1/v = 4/40 + 1/40

→ 1/v = (4+1)/40

→ 1/v = 5/40

→ v = 40/5

→ v = 8 cm

✭ Magnification :

→ M = -v/u

→ M = -8/-40

→ M = 0.2

∴ The image is virtual, erect and diminished

Answer 2

Given :-

Type of mirror = Convex

Radius of curvature = 20 cm

Object distance = -40 cm

To Find :-

Position and magnification of object

Solution :-

For finding position we will first find its focal length

$\sf \: focal \: length = \dfrac{radius}{2} = \dfrac{20}{2} = 10$

Now,

By using lens formula

$\sf \: \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}$

$\sf \: \dfrac{1}{v} + \dfrac{1}{( - 40)} = \dfrac{1}{10}$

$\sf \dfrac{1}{v} - \dfrac{1}{40} = \dfrac{1}{10}$

$\sf \: \dfrac{1}{v} = \dfrac{1}{40} + \dfrac{1}{10}$

$\sf \: \dfrac{1}{v} = \dfrac{4 + 1}{40}$

$\sf \: \dfrac{1}{v} = \dfrac{5}{40}$

$\sf \: v = \dfrac{40}{5}$

$\sf \: v = 8 \: cm$

Hence,

The position is 8 cm

Now

Let's find magnification

Magnification = -v/u

Magnification = -8/40

Magnification = 0.2