Question

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

Answer 1

Answer:B::C

Explanation:

Substituting ,u=-40cm and f=+40cm in the mirror formula (1/v)+(1/u)=(1/f),wehave(1/v)+(1/-40)=(1/+40),we≥t,v= +20 cmMagnification,m=-(v/u)=-(+20)/(-40)=+(1/2)Magnificationis+(1/2)`, which implies that image is erect, virtual and half in size.

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

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

Now,

• By using lens formula

⠀⠀ ⠀ ⠀⠀⠀⠀⠀⠀${\sf\color{Thistle}\dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}$

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

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

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

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

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

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

$\tt \: v = 8 \: cm$

Hence,

• The position is 8 cm

Now

• Let's find magnification

Magnification = -v/u

Magnification = -8/40

Magnification = 0.2