Question

an object 2cm high is placed 10cm in front of mirror . What type of mirror of mirror . What is radius of curvature is needed for a upright image is 4cm high ?

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Answer 1

Explanation:

It is a concave mirror.

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Answer 2

ANSWER:

Given:

• Height of object ($\sf H_o$ = 2cm
• Distance of object from mirror (u) = 10cm
• Height of image ($\sf H_i$ = 4cm

To Find:

• The type of mirror
• The radius of curvature (r)

Solution:

We are given that,

$\implies\sf Height\:of\: object (H_o)=2cm$

And,

$\implies\sf Height\:of\: image (H_i)=4cm$

We know that,

$\implies\sf Magnification=\dfrac{H_i}{H_o}$

So,

$\implies\sf Magnification=\dfrac{4\!\!\!/^{\:2}}{2\!\!\!/_{\:1}}$

Hence,

$\implies\sf Magnification=2 - - - -(1)$

We are given that,

$\implies\sf Distance\:of\: object (u)=10cm$

We also know that,

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

So,

$\implies\sf Magnification=\dfrac{-v}{-10}$

From (1),

$\implies\sf 2=\dfrac{v}{10}$

$\implies\sf v=20$

So,

$\implies\sf v=20 - - - -(2)$

We know that, by Mirror Formula,

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

Here, f is focal length, u is distance of object, and v is distance of image.

So,

$\implies\sf \dfrac{1}{f}=\dfrac{1}{20}+\dfrac{1}{-10}$

$\implies\sf \dfrac{1}{f}=\dfrac{1}{20}+\dfrac{-1}{10}$

$\implies\sf \dfrac{1}{f}=\dfrac{1-2}{20}$

$\implies\sf \dfrac{1}{f}=\dfrac{-1}{20}$

Hence,

$\implies\sf f=-20$

We know that,

$\implies\sf Radius\:of\: curvature =2f$

So,

$\implies\sf Radius\:of\: curvature =2(-20)$

Hence,

$\implies\sf Radius\:of\: curvature =-40cm$

As the radius is in negative, the mirror is a concave mirror.

And, radius is 40cm.