Question

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


[Note: Don't be greedy for points.]
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Answer 1

Answer:

Since image size is 1/3 of object size, so image distance is 1/3 of object distance because

h'/h=v/u

Using the mirror formula, we can calculate object distance and image distance

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

$\sf \: \frac{3}{u} - \frac{1}{u} = \frac{1}{20}$

$\sf \: \frac{3 - 1}{u} = \frac{1}{20}$

U=-40cm.

Using this, we can find image distance:

$\sf \boxed{ \sf\frac{1}{v} + \frac{1}{u} = \frac{1}{f} }$

$\sf \frac{1}{v} - \frac{1}{40} = \frac{1}{20}$

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

$\sf \: \frac{ - 2 + 1}{40} - \frac{1}{40}$

But above value of image distance does not match with our initial assumption. This means that the mirror is not a concave mirror but a convex mirror. Let us calculate with the assumptiom that it is a convex mirror.

$\sf \boxed{ \sf\frac{1}{v} + \frac{1}{u} = \frac{1}{f} }$

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

$\pink{ \sf \: \frac{1}{v} = - \frac{1}{20} + \frac{1}{40} - \frac{3}{40} }$

$\sf \: hence \: \: v \: = \frac{40}{3} cm$

Nature of mirror: convex mirror.

Nature of image: Smaller than object, erect and virtual.

Answer 2

An image smaller in size can be formed both by a concave mirror as well as convex mirror.

$\red{ \bf \: Case \: I: When \: mirror \: is \: concave, \: the \: mirror \: is \: real. }$

Hence,

• $\sf \: m = - \frac{1}{3}$
• $\sf f = - 20 \: m$

As,

$\sf \: m = - \frac{v}{u}$

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

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

★ Using Formula

$\sf \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

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

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

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

Now,

$\sf \: v = \frac{u}{3} = \frac{ - 80}{3} \: cm$

$\therefore $ Image is real and inverted.

$\rule{300px}{.3ex}$

$\red{ \bf \: Case \: II: When \: mirror \: is \: convex, the \: image \: is \: virtual. }$

• $\sf \: m = \frac{1}{3}$
• $\sf \: f = + 20 \: cm$

As,

$\sf \: m = - \frac{v}{u} = \frac{1}{3}$

$\sf \implies \: v = - \frac{u}{3}$

★ Using Formula,

$\sf \frac{1}{v} + \frac{1}{u} = \frac{1}{f}$

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

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

$\sf \: v = - \frac{( - 40)}{3} = \frac{40}{3} \: cm$

$\sf$

$\therefore $ Image is virtual and erect.

$\sf$

$\rule{300px}{.3ex}$