Question

An object 4.0 cm in length is placed at a distance of 10 cm in front of a convex

mirror of radius of curvature 20 cm. Find the position of the image, its nature

and size.​

Answer 1

$\huge\sf\pink{Answer}$

☞ The image is erect, virtual and enlarged

☞ v = 5 & m = 0.5

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$\huge\sf\blue{Given}$

✭ Height of the object = 4 cm

✭ u = -10 cm

✭ Radius of curvature = 20 cm

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$\huge\sf\gray{To \:Find}$

◈ Position, nature and size of the image?

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$\huge\sf\purple{Steps}$

So first we shall find the Focal Length,

➢ $\sf Focal \ Length = \dfrac{20}{2}$

➢ $\sf Focal \ Length = 10 \ cm$

Now as per mirror formula,

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

Substituting the values,

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

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

➝ $\sf \dfrac{1}{v}=\dfrac{1}{10}+\dfrac{1}{10}$

➝ $\sf \dfrac{1}{v} = \dfrac{2}{10}$

➝ $\sf \dfrac{1}{v} = \dfrac{1}{5}$

➝ $\sf \green{v = 5}$

Now magnification is given by,

$\underline{\boxed{\sf M = \dfrac{H_i}{H_o}}}$

Substituting the values,

➳ $\sf M = \dfrac{-v}{u}$

➳ $\sf M = \dfrac{-5}{-10}$

➳ $\sf M = \dfrac{1}{2}$

➳ $\sf \red{Magnification = 0.5}$

Now the height of the image will be,

»» $\sf M = \dfrac{H_i}{H_o}$

»» $\sf 0.5 = \dfrac{H_i}{4}$

»» $\sf 0.5\times 4 = H_i$

»» $\sf \orange{H_i = 2}$

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