Question

an object 3cm tall is placed at a distance of 40cm infront of a concave mirror whose focal Length is 20cm. what is the size of the image formed ??​

Answer 1

Answer :-

Size of the image formed is 3 cm [inverted].

Explanation :-

We have :-

→ Size of the object = 3 cm

→ Distance of the object = 40 cm

→ Focal length of the mirror = 20 cm

______________________________

Since, the mirror is concave :-

• f = -20 cm

• u = -40 cm

According to mirror formula :-

1/v + 1/u = 1/f

⇒ 1/v = 1/f - 1/u

⇒ 1/v = (-1)/20 - (-1)/40

⇒ 1/v = (-2 + 1)/40

⇒ 1/v = -1/40

⇒ -v = 40

⇒ v = -40 cm

______________________________

Now, by formula of magnification :-

m = -(v/u) = hᵢ/hₒ

⇒ -(v/u) = hᵢ/hₒ

⇒ -(-40/-40) = hᵢ/3

⇒ -1 = hᵢ/3

⇒ hᵢ = -3 cm

[Negative sign of 'hᵢ' shows that the image is inverted] .

Answer 2

Given:-

• An object 3cm tall is placed at a distance of 40cm infront of a concave mirror whose focal Length is 20cm.

To Find:-

• Size of the image formed

Formula Used:-

• ${\boxed{\bf{Mirror\: Formula: \dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}}$

• ${\boxed{\bf{Magnifcation\: Formula: m=\dfrac{-v}{u}=\dfrac{h'}{h}}}}$

where,

• f = focal length

• v = Distance of image

• u = Distance of object

• h = height of object

• h' = height of image

Solution:-

Using Mirror Formula,

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

Here,

• f = -20 cm

• u = -40 cm

Putting Values,

$\sf :\implies \dfrac{1}{-20}=\dfrac{1}{v}+\dfrac{1}{-40}$

$\sf :\implies \dfrac{1}{v}=\dfrac{1}{40}-\dfrac{1}{20}$

$\sf :\implies \dfrac{1}{v}=\dfrac{1-2}{40}$

$\sf :\implies \dfrac{1}{v}=\dfrac{-1}{40}$

$\bf :\implies v=-40\:cm$

Now, Using Magnification Formula,

$\sf :\implies\:m=\dfrac{-v}{u}=\dfrac{h'}{h}$

$\sf :\implies\:\dfrac{-(-40)}{-40}=\dfrac{h'}{3}$

$\sf :\implies\:\dfrac{-4}{4}=\dfrac{h'}{3}$

$\sf :\implies\:h'=3\times -1$

$\bf :\implies\:h'=-3\:cm$

{Here, Negative sign shows that the Image is Inverted.}

Hence, The size of the image formed is 3cm.

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