Question

An object of height 15 cm is placed at a distance of 25 cm in front of a convex mirror of focal length 12 cm. Find the position of the image and length of the image ​

Answer 1

$\large{\underline{\underline{\mathfrak{Answer :}}}}$

• Image distance is 8 cm
• Height of Image is 5 cm

$\rule{200}{0.5}$

$\underline{\underline{\mathfrak{Step-By-Step-Explanation :}}}$

Given :

• Height of object (Ho) = 15 cm
• Object Distance (u) = -25 cm
• Focal length (f) = 12 cm
• Convex mirror

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To Find :

• Position of Image
• Height of Image (Hi)

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Solution :

We have mirror formula :

$\large{\boxed{\sf{\dfrac{1}{f} \: = \: \dfrac{1}{v} \: + \: \dfrac{1}{u}}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{f} \: - \: \dfrac{1}{u}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{12} \: - \: \bigg( \dfrac{1}{-25} \bigg)}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{12} \: + \: \dfrac{1}{25}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{25 \: + \: 12}{300}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{37}{300}}} \\ \\ \implies {\sf{v \: = \: \dfrac{300}{37}}} \\ \\ \implies {\sf{v \: = \: 8.1}} \\ \\ \underline{\sf{\therefore \: Image \: distance \: is \: 8 \: cm}}$

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• Now, we have to find height of Image

Use formula :

$\large{\boxed{\sf{\dfrac{-v}{u} \: = \: \dfrac{H_i}{H_o}}}} \\ \\ \implies {\sf{\dfrac{-8}{-25} \: = \: \dfrac{H_i}{15}}} \\ \\ \implies {\sf{H_i \: = \: \dfrac{8 \: \times \: 15}{25}}} \\ \\ \implies {\sf{H_i \: = \: \dfrac{120}{25}}} \\ \\ \implies {\sf{H_i \: = \: 4.8}} \\ \\ \implies {\sf{H_i \: = \: 5}} \\ \\ \underline{\sf{\therefore \: Height \: of \: image \: is \: 5 \: cm}}$

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

★ Given :

Height of object (H) = 15 cm

Object Distance (u) = -25 cm

Focal length = 12 cm

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★ To Find :

We have to find tge position of the image and length of the image.

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★ Solution :

We know that,

$\Large{\implies{\boxed{\boxed{\sf{\frac{1}{f} = \frac{1}{v} + \frac{1}{u}}}}}}$

______________[Put Values]

$\sf{→ \frac{1}{12} = \frac{1}{v} + \frac{-1}{25}} \\ \\ \sf{→ \frac{1}{12} + \frac{1}{25}} \\ \\ \bf{Taking LCM} \\ \\ \sf{→ \frac{1}{v} = \frac{25 + 12}{300}} \\ \\ \sf{→ \frac{1}{v} = \frac{37}{300}} \\ \\ \sf{→ v = \frac{300}{37}} \\ \\ \bf{→ v = 8.10} \\ \\ \Large{\implies{\boxed{\boxed{\sf{ v = 8.10 \: cm}}}}}$

$\therefore$ Image Distance is 8.1 cm.

$\rule{200}{2}$

Now, We will find Magnification :

$\Large{\implies{\boxed{\boxed{\sf{M = \frac{-v}{u} = \frac{H'}{H}}}}}}$

$\sf{→ M = \frac{\cancel{-}8.1}{\cancel{-}25}} \\ \\ \sf{→ M = \frac{8.1}{25}} \\ \\ \bf{→ M = 0.324}$

$\rule{150}{2}$

$\bf{Now,} \\ \\ \sf{→ M = \frac{H'}{H}} \\ \\ \sf{→ 0.324 = \frac{H'}{15}} \\ \\ \sf{→ H' = 15 \times 0.24} \\ \\ \bf{→ H' = 4.86 \: cm} \\ \\ \Large{\implies{\boxed{\boxed{\sf{H' = 4.86 \: cm}}}}}$

$\therefore$ Image height is 4.86 cm.