Question

for concave mirror focal length is -3cm the object distance is -10cm and height of object h¹ is 2cm find possession , size and nature of image​

Answer 1

Given:-

• Focal length = -3 cm
• Object distance = -10 cm
• height of the object (h) = 2 cm

To Find:-

• Position of the image
• Size of the image
• Nature of the image

Solution:-

Applying mirror formula:-

$\dag\underline{\red{\boxed{\blue{\rm{\dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}}}}}$

Applying all the values in the formula:-

$= \sf{\dfrac{1}{-3} = \dfrac{1}{v} + \dfrac{1}{-10}}$

$= \sf{\dfrac{-1}{3} = \dfrac{1}{v} - \dfrac{1}{10}}$

$= \sf{\dfrac{-1}{3} + \dfrac{1}{10} = \dfrac{1}{v}}$

$= \sf{\dfrac{-10 + 3}{30} = \dfrac{1}{v}}$

$= \sf{\dfrac{-7}{30} = \dfrac{1}{v}}$

$= \sf{\dfrac{-30}{7} = v}$

$= \sf{v = -4.2\:cm}$

$\boxed{\underline{\red{\rm{\therefore\:The\:position\:of\:the\:image\:is\:at\:4.2\:cm\:(i.e.\:between\:f\:and\:c}}}}$

Now,

Applying magnification formula for mirrors:-

$\dag\underline{\blue{\boxed{\red{\rm{m = \dfrac{-v}{u} = \dfrac{h'}{h}}}}}}$

$= \sf{m = \dfrac{-(-4.2)}{-10} = \dfrac{h'}{2}}$

$\implies \sf{\dfrac{4.2\times 2}{-10} = h'}$

$= \sf{\dfrac{8.4}{-10} = h'}$

$= \sf{h' = -0.84\:cm}$

$\boxed{\underline{\orange{\rm{\therefore\:The\:height\:of\:the\:image\:is\:0.84\:cm}}}}$

Also, the magnification is negative.

Hence,

$\boxed{\underline{\green{\rm{The\:nature\:of\:the\:image\:is\:Real\:and\:Inverted}}}}$

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Points to remember!!!

• If the magnification of an image formed is negative, then the nature of the image will be real and inverted.
• If the magnification of an image is positive, then the nature of the image will be virtual and erect.
• h' is used to denote the height of the image.
• h is used to denote the height of the object.
• Object distance(u) is always taken as negative.

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