Question

An object 1 cm tall is placed 30 cm in front of diverging mirror of focal length 20 cm.

Find the size and position of the image formed by it.​

Answer 1

Answer:

Hey mate here is ur answer. Refer photo

Answer 2

GIVEN

An object 1 cm tall is placed 30 cm in front of diverging mirror of focal length 20 cm.

• Object height ( $\sf{h_o}$ ) = 1 cm
• Image distance (u) = -30 cm
• Focal length (f) = 20 cm

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TO FIND

The size and position of the image formed by the mirror.

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WE MUST KNOW

◙ Mirror formula :-

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

◙ Magnification :-

$\sf{\dfrac{h_i}{h_o}=\dfrac{-v}{u} }$

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SOLUTION

​Using the formula :-

$\displaystyle{\sf{\frac{1}{v}+\frac{1}{-30}=\frac{1}{20} }}\\\\\displaystyle{\sf{\longrightarrow \frac{1}{v}-\frac{1}{30}=\frac{1}{20} }}\\\\\displaystyle{\sf{\longrightarrow \frac{1}{v}=\frac{1}{20}+\frac{1}{30} }}\\\\\displaystyle{\sf{\longrightarrow \frac{1}{v}=\frac{3+2}{60} }}\\\\\displaystyle{\sf{\longrightarrow \frac{1}{v}=\frac{5}{60} }}\\\\\displaystyle{\sf{\longrightarrow \frac{1}{v}=\frac{1}{10} }}\\\\\boxed{\displaystyle{\sf{\longrightarrow v=12\ cm }}}$

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$\sf{Magnification = \dfrac{-v}{u} }$

$\sf{\longrightarrow Magnification=\dfrac{-12}{-30} }\\\\\sf{\longrightarrow Magnification=\dfrac{2}{5} }\\\\\boxed{\sf{\longrightarrow Magnification=0.4\ cm}}$

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Now,

$\sf{\dfrac{h_i}{h_o}=\dfrac{-v}{u} }\\\\\sf{\longrightarrow \dfrac{h_i}{1}=\dfrac{-12}{-30} }\\\\\sf{\longrightarrow \dfrac{h_i}{1}=\dfrac{-12}{-30} }\\\\\sf{\longrightarrow h_i=\dfrac{12}{30} }\\\\\sf{\longrightarrow h_i=\dfrac{2}{3} }\\\\\boxed{\sf{\longrightarrow h_i=1.5\ cm}}$

CONCLUSION :-

• The image is 12 cm on the right side of the convex mirror.
• The image is a virtual and erect image, as the magnification is positive.
• The image size (h$\sf{_i}$) is greater than the object size (h$\sf{_o}$), so the image if magnified.