Question

$\huge\underline\mathfrak\blue{Question-}$

An object of size 7 cm is placed at 27 cm in front of a concave mirror of focal length is 18 cm. At what distance from the mirror should a screen be placed so that a sharp focused image can be obtained?

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Answer 1

Answer:

yahi hai mera answers

Explanation:

aapko Sahi lage to thanks jarru bhejna

Answer 2

$\large{\underline{\underline{\mathfrak{\sf{\red{Answer-}}}}}}$

Screen should be placed at a distance of 54 cm from the concave mirror so that a sharp and focused image can be formed.

$\large{\underline{\underline{\mathfrak{\sf{\red{Explanation-}}}}}}$

$\begin{lgathered}\bold{Given} \begin{cases}\sf{Focal\:length(f)=-18\:cm} \\ \sf{Distance\:of\:object(u)=-27\:cm}\\ \sf{height\:of\:object=7\:cm}\end{cases}\end{lgathered}$

$\large{\boxed{\mathfrak{\sf{\green{To\:find-}}}}}$

• Distance of object ( u ).

$\large{\underline{\boxed{\mathfrak{\sf{\green{Formula\:used-}}}}}}$

• Mirror formula

$\implies$ $\large{\underline{\boxed{\sf{\pink{\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}}}}$

• Linear magnification ( m )

$\implies$ $\large{\underline{\boxed{\sf{\blue{\dfrac{h'}{h}=\dfrac{-v}{u}}}}}}$

$\large{\boxed{\mathfrak{\sf{\green{Solution-}}}}}$

By using the mirror formula,

★ $\large{\boxed{\sf{\orange{\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u}}}}}$

$\leadsto$ $\sf{\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}}$

Putting the given values,

$\leadsto$ $\sf{\dfrac{1}{v}=\dfrac{1}{-18}-\dfrac{1}{-27}}$

$\leadsto$ $\sf{\dfrac{1}{v}=\dfrac{1}{-18}+\dfrac{1}{27}}$

By taking LCM,

$\leadsto$ $\sf{\dfrac{1}{v}=\dfrac{-3+2}{54}}$

$\leadsto$ $\sf{\dfrac{1}{v}=\dfrac{-1}{54}}$

We get,

$\large{\boxed{\mathfrak{\purple{\sf{v=-54\:cm}}}}}$

$\therefore$ Screen should be placed at a distance of 54 cm from the concave mirror so that a sharp and focused image can be formed.

$\rule{200}2$

Linear magnification :

$\implies$ $\large{\underline{\boxed{\sf{\blue{\dfrac{h'}{h}=\dfrac{-v}{u}}}}}}$

Put the given values in the above formula,

$\leadsto$ $\sf{\dfrac{h'}{7}=\cancel{\dfrac{-(-54)}{-27}}}$

$\leadsto$ $\sf{\dfrac{h'}{7}=-2}$

$\leadsto$ $\sf{h'=-2×7}$

we get,

$\large{\boxed{\mathfrak{\purple{\sf{h'=-14\:cm}}}}}$

Characteristics of image :

• Real
• Inverted
• Larger than object
• Placed at a distance of 54 cm

$\rule{200}2$

Points you should know :

• A mirror whose reflecting surface faces towards the centre of that sphere of which a mirror is a part is called a Concave mirror.

• It is having negative focal length.