Question

Size of image of an object formed by a mirror having a focal length of 20 cm, is observed to
be reduced to 1/3 of its size. At what distance the object has been placed from the mirror?
What is the nature of image and the mirror? ​

Answer 1

Answer:

HII!

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lets see the answer in cases!1

CASE 1:-  m=-1/3(in case of concave mirror)

f= -20 cm

m= -1/3 =-v/u

v=u/3

1/f= 1/v +1/u .

=> -1/20 = 3/u + 1/ u= 4/u

therefore.,, u = -80 cm.

CASE 2:-

M=+1/3 (IN CASE OF CONVEX MIRROR)

f=+20 cm.

m=+1/3 =-v/u

v=-u/3.

i/f=1/v+1/u.

=> 1/20= -3/4 + 1/4 =-2/4

therefore, u= -40 cm.

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Answer 2

$\red\bigstar$Calculation:-

An image smaller in size can be formed by both by a concave mirror as well as convex mirror.Here, the two cases arises .So let's see all two cases .

Case 1:- When mirror is concave ,the image is real.

Hence,

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$\sf \: \: \mapsto Magnification, m=- \dfrac{1}{3} , \\ \: \sf \mapsto Focal \: length,f=-20cm$

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As we know the expression of linear magnification,

$\underline{ \boxed{ \bullet \mapsto\sf \: m = - \frac{v}{u} }}$

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$\sf \: \: \therefore - \dfrac{1}{3} = - \dfrac{v}{u}$

$\sf\implies \: v = \dfrac{u}{3}$

Now, using mirror formula,

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$\underline{ \boxed { \sf \bullet \mapsto \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} } }$

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$\implies \sf \dfrac{3}{u} + \dfrac{1}{u} = \dfrac{ \: 1 \: }{ - 20 \: }$

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$\implies \sf \dfrac{4}{u} = \dfrac{1}{ - 20}$

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$\implies \sf \: u = - 80 \: cm$

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$\sf Now,v= \dfrac{u}{3} = \dfrac{-80}{2}cm$

Therefore,Image is real and inverted.

Case-2:- When mirror is convex.

$\sf \: \: \mapsto Magnification, m= \dfrac{1}{3} , \\ \: \sf \mapsto Focal \: length,f= + 20cm$

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Again using linear expression of magnification

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$\underline{ \boxed{ \bullet \mapsto\sf \: m = - \frac{v}{u} }}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

$\sf \: \: \therefore \dfrac{1}{3} = - \dfrac{v}{u}$

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$\implies \sf v = - \dfrac{u}{3}$

Again using mirror formula

$\underline{ \boxed { \sf \bullet \mapsto \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f} } }$

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$\implies \sf - \dfrac{3}{u} + \dfrac{1}{u} = \dfrac{1}{20}$

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$\implies \sf \: u = - 40 \: cm$

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$\sf Now,v= -\dfrac{-40}{3} = \dfrac{40}{3} cm$

Therefore ,In case 2: The image is virtual and erect.