Question

an object of height 2 cm is placed at a distance of 20 cm in front of a concave of focal length 12 cm. find the position, size and nature of the image formed.

Answer 1

GivEn:

• Height of an object $\sf h_{o}$ = 2 cm

• Object distance, u = - 20cm

• Focal length = - 12 cm

To find:

• position, size and nature of the image formed.

SoluTion:

Here, Mirror is concave so,

★ Focal length of object is -ve.

★ And we know that, object distance is always taken as -ve.

$\bf \underline{\bigstar\;As\; we \;know\; that,}$

${\boxed{\sf{Mirror\;Formula = \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}}$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf \dfrac{1}{- 12} = \dfrac{1}{v} + \dfrac{1}{- 20}$

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$:\implies\sf \dfrac{1}{v} = \dfrac{1}{- 12} - \dfrac{1}{-20}$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf \dfrac{1}{v} = - \dfrac{1}{12} + \dfrac{1}{20}$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf \dfrac{1}{v} = \dfrac{ - 5 + 3}{60}$

⠀⠀⠀⠀⠀⠀⠀

$:\implies\sf \dfrac{1}{v} = \cancel{ \dfrac{ - 2}{60}}$

⠀⠀⠀⠀⠀⠀⠀

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

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$:\implies{\underline{\boxed{\sf{\pink{v = -30\;cm}}}}}\;\bigstar$

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Now,

We have to find Magnification,

So, As we know that,

${\boxed{\sf{Magnification, m = \dfrac{h_{\;(i)}}{h_{\;(o)}} = \dfrac{-v}{u}}}}$

$:\implies\sf m = \cancel{ \dfrac{30}{- 20}}$

$:\implies{\underline{\boxed{\sf{\purple{m = - 1.5}}}}}\;\bigstar$

Now, we can find height of image, $\sf h_{\;(i)}$

$:\implies\sf m = \dfrac{h_{\;(i)}}{h_{\;(o)}}$

$:\implies\sf - 1.5 = \dfrac{h_{\;(i)}}{2}$

$:\implies\sf h_{\;(i)} = - 1.5 \times 2$

$:\implies{\underline{\boxed{\sf{\blue{h_{\;(i)} = - 3\;cm}}}}}\;\bigstar$

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$\therefore$ The nature of image is,

• Real
• Inverted
• Magnified

Answer 2

Solution :

$\underline{\bf{Given\::}}}$

• Distance of object from lens, (u) = -20 cm
• Height of object, (h) = 2 cm
• Focal length, (f) = -12 cm

$\underline{\bf{Explanation\::}}}$

As we know that formula of the lens;

$\boxed{\bf{\dfrac{1}{f} =\dfrac{1}{v} -\dfrac{1}{u} }}}$

A/q

$\mapsto\sf{\dfrac{1}{f} =\dfrac{1}{v} -\dfrac{1}{u} }\\\\\\\mapsto\sf{\dfrac{1}{-12} =\dfrac{1}{v} -\dfrac{1}{(-20)} }\\\\\\\mapsto\sf{\dfrac{1}{-12} =\dfrac{1}{v} +\dfrac{1}{20} }\\\\\\\mapsto\sf{\dfrac{1}{v} =\dfrac{1}{-12} -\dfrac{1}{20} }\\\\\\\mapsto\sf{\dfrac{1}{v} =\dfrac{-5-3}{60} }\\\\\\\mapsto\sf{\dfrac{1}{v} =\dfrac{-8}{60} }\\\\\\\mapsto\sf{-8v =60}\\\\\mapsto\sf{-v=\cancel{60/8}}\\\\\mapsto\sf{-v=7.5\:cm}\\\\\mapsto\bf{v=-7.5\:cm}$

So, Image will be formed on the same side of object at a distance of 7.5 cm from the lens.

Now;

Using formula of the linear magnification :

$\mapsto\bf{m=\dfrac{Height\:of\:image\:(I)}{Height\:of\:object\:(O)} =\dfrac{Distance\:of\:Image}{Distance\:of\:object} =\dfrac{v}{u} }$

$\mapsto\sf{m=\dfrac{h_2}{h_1} =\dfrac{v}{u} }\\\\\\\mapsto\sf{\dfrac{h_2}{2} =\dfrac{-7.5}{-20}} \\\\\\\mapsto\sf{-20h_2=-7.5\times 2}\\\\\mapsto\sf{-20h_2= -15}\\\\\mapsto\sf{h_2 = \cancel{-15/-20}}\\\\\mapsto\bf{h_2 =0.75\:cm}$

∴ Height of image will be 0.75 cm (erect) .