Question

An object of 10 cm height is placed in front of a convex mirror of focal length 20 cm at a distance of 30 cm from mirror. find the height of the mirror.​

Answer 1

$\large{\red{\frak{ Given }}}\begin{cases} \textsf{ Height of the object is \textbf{10 cm}.}\\\textsf{ Focal length of the convex mirror is \textbf{20 cm}.}\\\textsf{ The object distance is \textbf{ 30 cm}.}\end{cases}$

$\large{\red{\frak{ To \ Find }}}\begin{cases} \textsf{ The image distance .}\\\textsf{ The height of the image .}\end{cases}$

Given that the height of the object is 10cm placed at a distance of 30 cm from the mirror of focal length 20 cm . Here we can use the mirror formula to find the image distance .

$\sf:\implies\pink{ \dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}}\\\\\sf:\implies \dfrac{1}{v}+\dfrac{1}{-30cm}=\dfrac{1}{20cm} \\\\\sf:\implies \dfrac{1}{v}=\dfrac{1}{30cm}+\dfrac{1}{20cm}\\\\\sf:\implies \dfrac{1}{v}=\dfrac{3+2}{60cm }\\\\\sf:\implies \dfrac{1}{v}=\dfrac{5}{60cm }\\\\\sf:\implies \dfrac{1}{v}=\dfrac{1}{12 cm}\\\\\sf:\implies\boxed{\pink{\frak{ Image \ Distance = 12 cm }}}$

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$\purple{\bigstar}\underline{\boldsymbol{ According\ to \ the \ Question :- }}$

$\sf:\implies \pink{ Magnification (m)=\dfrac{Height _{(image)}}{Height_{(object)}}=\dfrac{Distance_{(image)}}{Distance_{(object)}}}\\\\\sf:\implies \dfrac{h_i}{h_o}=\dfrac{v}{u}\\\\\sf:\implies \dfrac{h_i}{10 cm }=\dfrac{12cm}{30cm}\\\\\sf:\implies h_i =\dfrac{12}{30}\times 10 cm \\\\\sf:\implies \underset{\blue{\sf Required \ Height }}{\underbrace{\boxed{\pink{\frak{ Height_{(image)} = 4 cm }}} }}$

Here we got the image height as 4cm . And the object height was 10 cm . This implies that the height of the image is less than the height of the object . When in case of ray diagram when an object is placed in front of a convex mirror then an diminished virtual image is formed . All these characteristics matches with our answer . Hence our answer is correct .

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Answer 2

Answer:

The image will be at a distance of 12 cm behind the mirror . Therefore, the height of the image is 0.4 cm.