Question

An object 5 cm in length is placed at a distance of 30 cm in front of a convex mirror of radius of curvature 40 cm. Find the position of the image, its nature and size.

Answer 1

★$\huge\bold{\underline{\rm{\red{Question :-}}}}$

• An object 5 cm in length is placed at a distance of 30 cm in front of a convex mirror of radius of curvature 40 cm. Find the position of the image, its nature and size.

★$\huge\bold{\underline{\rm{\pink{Answer:-}}}}$

• The image is formed at a distance of 12cm behind the convex mirror.
• Height of an image is 2cm.

★$\large\bold{\underline{\rm{\blue{Given :-}}}}$

• $\small\bold{\sf{u=-30cm}}$
• $\small\bold{\sf{h=5cm}}$
• $\small\bold{\sf{R=40cm...(covex\:mirror)}}$

★${\underline{According \: to \: the \: question:-}}$

$\sf \: f = \dfrac{R}{2} \\ \sf= \frac{40}{2} \\ = \sf + 20cm...(convex \: mirror)$

From the mirror formula we get,

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

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

$\implies\sf \: \dfrac{1}{v} = \dfrac{1}{ +20} - \dfrac{1}{ - 30}$

$\implies\sf \: \dfrac{1}{v} = \dfrac{ - 30 - ( + 20)}{20 \times - 30} \\ \implies\sf \: \frac{1}{v} = \frac{ - 30 - 20}{ - 600} \\ \implies\sf \: \frac{1}{v} = \frac{ - 50}{ - 600} \\ \implies\sf \:v = 12cm$

The image is formed at a distance of 12cm behind the convex mirror. Hence, it is virtual and erect.

$\implies\sf \: m = - \dfrac{v}{u} \\ \implies\sf \: m = - \frac{12}{ - 30}\\ \implies\sf \: m = 0.4cm$

Also,

$\implies\sf \: m = \dfrac{h_{i} }{h _{0}} \\ \implies\sf \: 0.4 = \frac{h _{i} }{5} \\ \implies\sf \: h _{i} = 5 \times 0.4 = 2cm$

Thus, height of an image is 2cm.

Therefore,we can say that the image is diminished.

Answer 2

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Question}}}}}}}}$

An object 5 cm in length is placed at a distance of 30 cm in front of a convex mirror of radius of curvature 40 cm. Find the position of the image, its nature and size.

${\huge{\bold{\purple{\bigstar{\boxed{\boxed{\bf{Solution}}}}}}}}$

${\bold{\blue{\boxed{\bf{Given }}}}}$

• it is an convex mirror
• u = - 30cm
• h = 5cm
• r = 40 cm

${\bold{\blue{\boxed{\bf{find }}}}}$

• position of the image
• its size
• its nature

${\bold{\blue{\boxed{\bf{Solution}}}}}$

${\bold{\green{\boxed{\bf{focal\:length= \dfrac{R}{2} }}}}}$

focal length = 40/2 = 20cm

${\bold{\green{\boxed{\bf{\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v} }}}}}$

$\sf\longrightarrow\dfrac{1}{20} = \dfrac{1}{-30}+ \dfrac{1}{v}$

$\sf\longrightarrow\dfrac{1}{v} = \dfrac{1}{20}- \dfrac{-1}{30}$

$\sf\longrightarrow\dfrac{1}{v} = \dfrac{30+20}{600}$

$\sf\longrightarrow\dfrac{1}{v} = \dfrac{50}{600}$

$\sf\longrightarrow\dfrac{1}{v} = \dfrac{1}{12}$

$\sf\longrightarrow v = 12$

${\bold{\green{\boxed{\bf{m = \dfrac{h_i}{h_0} }}}}}$--------(i)

${\bold{\green{\boxed{\bf{m = \dfrac{-v}{u} }}}}}$-----------(ii)

therefore from eq (i) and (ii)

${\bold{\green{\boxed{\bf{\dfrac{-v}{u}= \dfrac{h_i}{h_0} }}}}}$

$\sf\longrightarrow{\dfrac{-12}{-30}= \dfrac{h_i}{5} }$

$\sf\longrightarrow{\dfrac{-12\times 5}{-30}= {h_i} }$

$\sf\longrightarrow h_i = 2$

${\bold{\pink{\boxed{\bf{\dag position = 12cm\: behind \:the \:mirror }}}}}$

${\bold{\pink{\boxed{\bf{\dag size= 2cm }}}}}$

${\bold{\pink{\boxed{\bf{\dag nature = diminished \: virtual \: erect }}}}}$

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THANK YOU❤